Importance of differentiationWhy is the derivative important?What is the use of Calculus?what is the use of...

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Importance of differentiation


Why is the derivative important?What is the use of Calculus?what is the use of derivativesWhy do we differentiate?What do instantaneous rates of change really represent?Some basic practical applications of CalculusUnexpected Practical Applications of CalculusApplications of derivatives outside mathematics and physicsWhat should the high school math curriculum consist of?What is the importance of Calculus in today's Mathematics?What are the practical applications of the Taylor Series?What makes a good mathematician?When should I start learning Set Theory?What exactly are limits used for in real life applications?Application of differential calculusCalculus & Analytic Geometry VS Vector CalculusHow important are differentiation and integration tricks and techniques to mathematicians?Why is the derivative important?













2












$begingroup$


I have just started learning about differentiation. I know that differentiation is about finding the slopes of curves of functions and etc.

I have many saying that differential and integral calculus are important tools of math and have many real-life applications.



So I want to how differentiation, which about finding slopes of functions, is important in real life. Explain it with some applications.



I am sorry if this is duplicate of other question. But as far as I have searched, I have not found an answer to this question










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Possible duplicate of Applications of derivatives outside mathematics and physics
    $endgroup$
    – Don Thousand
    2 hours ago










  • $begingroup$
    @DonThousand I don't think it is a duplicate of that question. What that question asks is about application outside maths and physics. But I am including maths and physics. So, I feel that there is a difference...
    $endgroup$
    – rash
    2 hours ago










  • $begingroup$
    Sure, but that question definitely answers this.
    $endgroup$
    – Don Thousand
    2 hours ago










  • $begingroup$
    The answer to this kind of question is dynamic. Who knows maybe I will a different variety of answers @DonThousand
    $endgroup$
    – rash
    2 hours ago










  • $begingroup$
    The derivative is the instantaneous rate of change of a function. So whenever we want discuss the rate of change of some quantity, which is all the time in physics or engineering, we find ourselves talking about derivatives.
    $endgroup$
    – littleO
    1 hour ago
















2












$begingroup$


I have just started learning about differentiation. I know that differentiation is about finding the slopes of curves of functions and etc.

I have many saying that differential and integral calculus are important tools of math and have many real-life applications.



So I want to how differentiation, which about finding slopes of functions, is important in real life. Explain it with some applications.



I am sorry if this is duplicate of other question. But as far as I have searched, I have not found an answer to this question










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Possible duplicate of Applications of derivatives outside mathematics and physics
    $endgroup$
    – Don Thousand
    2 hours ago










  • $begingroup$
    @DonThousand I don't think it is a duplicate of that question. What that question asks is about application outside maths and physics. But I am including maths and physics. So, I feel that there is a difference...
    $endgroup$
    – rash
    2 hours ago










  • $begingroup$
    Sure, but that question definitely answers this.
    $endgroup$
    – Don Thousand
    2 hours ago










  • $begingroup$
    The answer to this kind of question is dynamic. Who knows maybe I will a different variety of answers @DonThousand
    $endgroup$
    – rash
    2 hours ago










  • $begingroup$
    The derivative is the instantaneous rate of change of a function. So whenever we want discuss the rate of change of some quantity, which is all the time in physics or engineering, we find ourselves talking about derivatives.
    $endgroup$
    – littleO
    1 hour ago














2












2








2


1



$begingroup$


I have just started learning about differentiation. I know that differentiation is about finding the slopes of curves of functions and etc.

I have many saying that differential and integral calculus are important tools of math and have many real-life applications.



So I want to how differentiation, which about finding slopes of functions, is important in real life. Explain it with some applications.



I am sorry if this is duplicate of other question. But as far as I have searched, I have not found an answer to this question










share|cite|improve this question











$endgroup$




I have just started learning about differentiation. I know that differentiation is about finding the slopes of curves of functions and etc.

I have many saying that differential and integral calculus are important tools of math and have many real-life applications.



So I want to how differentiation, which about finding slopes of functions, is important in real life. Explain it with some applications.



I am sorry if this is duplicate of other question. But as far as I have searched, I have not found an answer to this question







calculus soft-question education






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago







rash

















asked 2 hours ago









rashrash

44312




44312








  • 3




    $begingroup$
    Possible duplicate of Applications of derivatives outside mathematics and physics
    $endgroup$
    – Don Thousand
    2 hours ago










  • $begingroup$
    @DonThousand I don't think it is a duplicate of that question. What that question asks is about application outside maths and physics. But I am including maths and physics. So, I feel that there is a difference...
    $endgroup$
    – rash
    2 hours ago










  • $begingroup$
    Sure, but that question definitely answers this.
    $endgroup$
    – Don Thousand
    2 hours ago










  • $begingroup$
    The answer to this kind of question is dynamic. Who knows maybe I will a different variety of answers @DonThousand
    $endgroup$
    – rash
    2 hours ago










  • $begingroup$
    The derivative is the instantaneous rate of change of a function. So whenever we want discuss the rate of change of some quantity, which is all the time in physics or engineering, we find ourselves talking about derivatives.
    $endgroup$
    – littleO
    1 hour ago














  • 3




    $begingroup$
    Possible duplicate of Applications of derivatives outside mathematics and physics
    $endgroup$
    – Don Thousand
    2 hours ago










  • $begingroup$
    @DonThousand I don't think it is a duplicate of that question. What that question asks is about application outside maths and physics. But I am including maths and physics. So, I feel that there is a difference...
    $endgroup$
    – rash
    2 hours ago










  • $begingroup$
    Sure, but that question definitely answers this.
    $endgroup$
    – Don Thousand
    2 hours ago










  • $begingroup$
    The answer to this kind of question is dynamic. Who knows maybe I will a different variety of answers @DonThousand
    $endgroup$
    – rash
    2 hours ago










  • $begingroup$
    The derivative is the instantaneous rate of change of a function. So whenever we want discuss the rate of change of some quantity, which is all the time in physics or engineering, we find ourselves talking about derivatives.
    $endgroup$
    – littleO
    1 hour ago








3




3




$begingroup$
Possible duplicate of Applications of derivatives outside mathematics and physics
$endgroup$
– Don Thousand
2 hours ago




$begingroup$
Possible duplicate of Applications of derivatives outside mathematics and physics
$endgroup$
– Don Thousand
2 hours ago












$begingroup$
@DonThousand I don't think it is a duplicate of that question. What that question asks is about application outside maths and physics. But I am including maths and physics. So, I feel that there is a difference...
$endgroup$
– rash
2 hours ago




$begingroup$
@DonThousand I don't think it is a duplicate of that question. What that question asks is about application outside maths and physics. But I am including maths and physics. So, I feel that there is a difference...
$endgroup$
– rash
2 hours ago












$begingroup$
Sure, but that question definitely answers this.
$endgroup$
– Don Thousand
2 hours ago




$begingroup$
Sure, but that question definitely answers this.
$endgroup$
– Don Thousand
2 hours ago












$begingroup$
The answer to this kind of question is dynamic. Who knows maybe I will a different variety of answers @DonThousand
$endgroup$
– rash
2 hours ago




$begingroup$
The answer to this kind of question is dynamic. Who knows maybe I will a different variety of answers @DonThousand
$endgroup$
– rash
2 hours ago












$begingroup$
The derivative is the instantaneous rate of change of a function. So whenever we want discuss the rate of change of some quantity, which is all the time in physics or engineering, we find ourselves talking about derivatives.
$endgroup$
– littleO
1 hour ago




$begingroup$
The derivative is the instantaneous rate of change of a function. So whenever we want discuss the rate of change of some quantity, which is all the time in physics or engineering, we find ourselves talking about derivatives.
$endgroup$
– littleO
1 hour ago










4 Answers
4






active

oldest

votes


















2












$begingroup$

The first real world application that comes into mind to me is calculating your "instantaneous speed", or perhaps more accurately, calculating your average speed over an infinitesimally small amount of time.



I'm sure you know that your average speed is given by $frac{text{distance traveled}}{text{time traveled}}$. So, what differentiation gives you is the ability to make the "time traveled" quantity very very small (tending to zero), giving you your speed at a given instant.



If you think about it, this is quite useful, especially if your'e driving a car and would like to know your current speed.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In general, the instantaneous rate of change in any quantity that varies continuously (or close enough to continuously) over time
    $endgroup$
    – Zubin Mukerjee
    2 hours ago



















2












$begingroup$

An important application
is finding the
extreme values
(min or max) or a function.



An example:
To find the angle
at which a thrown ball
goes the farthest,
derive the formula
for the distance traveled
as a function of the angle
and then differentiate it
and find the angle
which makes the derivative zero.
That is the angle with
maximum distance.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Differentiation of any function gives you the slope of the tangent at that point. Why? Consider a line with the equation:
    $$y=mx+c$$
    Consider two points $A(x,y)$ and another point that is slightly far from A, $B(x+triangle x,y + triangle y)$.
    ![enter image description here
    where $triangle y$ is the change in the function value for a change corresponding to $triangle x$. In other words, the change in $y$ occurs due to a change in $x$.
    $$y=mx+c$$
    $$triangle y=triangle {mx +c}=triangle {(mx)} + triangle c= mtriangle x$$
    So, now we calculate the slope of the line $AB$, the coordinates are $A(x,y)$ and $B(x+triangle x,y+triangle y)$. The slope is :
    $$frac{triangle y}{triangle x}=frac{mtriangle x}{triangle x}=m$$
    If you make the changes $triangle x$ and $triangle y$, very small tending to $0$ then the values change to $dy$ and $dx$. The slope is :
    $$lim_{triangle xrightarrow 0}frac{triangle y}{triangle x}=frac{dy}{dx}=m$$The visualization is like you bring the coordinate $B$ closer and closer to $A$ and keep on calculating slope until the point $B$ is so close that it coincides with the point $A$. Now you get the slope of the line joining $A$ and $A$ or in other words slope of the line at point $A$.



    Same things can be generalized for a function $f(x)$
    enter image description here
    Consider a function $f(x)$ and a point $A(x,y)$ on it. We take another point $B(x+triangle x,y+triangle y)$. Now we calculate the slope again of the line $AB$ as:
    $$frac{f(x+triangle x)-f(x)}{x+triangle x -x}=frac{f(x+triangle x)-f(x)}{triangle x}$$



    Now, the line $AB$ is actually a secant to the curve $f(x)$. Just bring the point B closer and closer to A that is keep decreasing $triangle x$ so that it becomes very small. Finally as $triangle x rightarrow 0$, the slope changes to $frac{dy}{dx}$ and the line $AB$ which was a secant back then has turned into a line $AA$ which means that it is a tangent at point $A$. That is why the slope $frac{dy}{dx}$ gives the slope of the tangent at any point.
    Finally,
    $$lim_{triangle x rightarrow 0 }frac{f(x+triangle)-f(x)}{triangle x}=frac{dy}{dx}$$



    This is also known as the first principle or Ab Initio method to compute the derivative.



    What else can you do once you know the slope of the line using differentiation. Anything and literally everything. Knowing the slopes of line will tell you the change in the slope is positive or negative giving the idea whether the function itself is increasing or decreasing function. Also if the slope of a tangent becomes $0$ at a point then there is a possibility of the maxima or minima at the same point. You can find the rate of change of quantity with respect to another like velocity or acceleration. The underlying concept is only one, the slope of the tangent.
    Hope this helps.....






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Derivatives are important in modeling physical phenomena via differential equations. These equations can model real-world processes much more accurately than a purely algebraic one. For example, the equation $theta’’+frac g l sintheta=0$ models the motion of a simple pendulum, where $theta$ is the angular displacement, $l$ is the length of the pendulum, and $g$ is the acceleration due to gravity.






      share|cite









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        Your Answer





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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        The first real world application that comes into mind to me is calculating your "instantaneous speed", or perhaps more accurately, calculating your average speed over an infinitesimally small amount of time.



        I'm sure you know that your average speed is given by $frac{text{distance traveled}}{text{time traveled}}$. So, what differentiation gives you is the ability to make the "time traveled" quantity very very small (tending to zero), giving you your speed at a given instant.



        If you think about it, this is quite useful, especially if your'e driving a car and would like to know your current speed.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          In general, the instantaneous rate of change in any quantity that varies continuously (or close enough to continuously) over time
          $endgroup$
          – Zubin Mukerjee
          2 hours ago
















        2












        $begingroup$

        The first real world application that comes into mind to me is calculating your "instantaneous speed", or perhaps more accurately, calculating your average speed over an infinitesimally small amount of time.



        I'm sure you know that your average speed is given by $frac{text{distance traveled}}{text{time traveled}}$. So, what differentiation gives you is the ability to make the "time traveled" quantity very very small (tending to zero), giving you your speed at a given instant.



        If you think about it, this is quite useful, especially if your'e driving a car and would like to know your current speed.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          In general, the instantaneous rate of change in any quantity that varies continuously (or close enough to continuously) over time
          $endgroup$
          – Zubin Mukerjee
          2 hours ago














        2












        2








        2





        $begingroup$

        The first real world application that comes into mind to me is calculating your "instantaneous speed", or perhaps more accurately, calculating your average speed over an infinitesimally small amount of time.



        I'm sure you know that your average speed is given by $frac{text{distance traveled}}{text{time traveled}}$. So, what differentiation gives you is the ability to make the "time traveled" quantity very very small (tending to zero), giving you your speed at a given instant.



        If you think about it, this is quite useful, especially if your'e driving a car and would like to know your current speed.






        share|cite|improve this answer









        $endgroup$



        The first real world application that comes into mind to me is calculating your "instantaneous speed", or perhaps more accurately, calculating your average speed over an infinitesimally small amount of time.



        I'm sure you know that your average speed is given by $frac{text{distance traveled}}{text{time traveled}}$. So, what differentiation gives you is the ability to make the "time traveled" quantity very very small (tending to zero), giving you your speed at a given instant.



        If you think about it, this is quite useful, especially if your'e driving a car and would like to know your current speed.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 hours ago









        Math Enthusiast Math Enthusiast

        455




        455












        • $begingroup$
          In general, the instantaneous rate of change in any quantity that varies continuously (or close enough to continuously) over time
          $endgroup$
          – Zubin Mukerjee
          2 hours ago


















        • $begingroup$
          In general, the instantaneous rate of change in any quantity that varies continuously (or close enough to continuously) over time
          $endgroup$
          – Zubin Mukerjee
          2 hours ago
















        $begingroup$
        In general, the instantaneous rate of change in any quantity that varies continuously (or close enough to continuously) over time
        $endgroup$
        – Zubin Mukerjee
        2 hours ago




        $begingroup$
        In general, the instantaneous rate of change in any quantity that varies continuously (or close enough to continuously) over time
        $endgroup$
        – Zubin Mukerjee
        2 hours ago











        2












        $begingroup$

        An important application
        is finding the
        extreme values
        (min or max) or a function.



        An example:
        To find the angle
        at which a thrown ball
        goes the farthest,
        derive the formula
        for the distance traveled
        as a function of the angle
        and then differentiate it
        and find the angle
        which makes the derivative zero.
        That is the angle with
        maximum distance.






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          An important application
          is finding the
          extreme values
          (min or max) or a function.



          An example:
          To find the angle
          at which a thrown ball
          goes the farthest,
          derive the formula
          for the distance traveled
          as a function of the angle
          and then differentiate it
          and find the angle
          which makes the derivative zero.
          That is the angle with
          maximum distance.






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            An important application
            is finding the
            extreme values
            (min or max) or a function.



            An example:
            To find the angle
            at which a thrown ball
            goes the farthest,
            derive the formula
            for the distance traveled
            as a function of the angle
            and then differentiate it
            and find the angle
            which makes the derivative zero.
            That is the angle with
            maximum distance.






            share|cite|improve this answer









            $endgroup$



            An important application
            is finding the
            extreme values
            (min or max) or a function.



            An example:
            To find the angle
            at which a thrown ball
            goes the farthest,
            derive the formula
            for the distance traveled
            as a function of the angle
            and then differentiate it
            and find the angle
            which makes the derivative zero.
            That is the angle with
            maximum distance.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 hours ago









            marty cohenmarty cohen

            74.3k549128




            74.3k549128























                2












                $begingroup$

                Differentiation of any function gives you the slope of the tangent at that point. Why? Consider a line with the equation:
                $$y=mx+c$$
                Consider two points $A(x,y)$ and another point that is slightly far from A, $B(x+triangle x,y + triangle y)$.
                ![enter image description here
                where $triangle y$ is the change in the function value for a change corresponding to $triangle x$. In other words, the change in $y$ occurs due to a change in $x$.
                $$y=mx+c$$
                $$triangle y=triangle {mx +c}=triangle {(mx)} + triangle c= mtriangle x$$
                So, now we calculate the slope of the line $AB$, the coordinates are $A(x,y)$ and $B(x+triangle x,y+triangle y)$. The slope is :
                $$frac{triangle y}{triangle x}=frac{mtriangle x}{triangle x}=m$$
                If you make the changes $triangle x$ and $triangle y$, very small tending to $0$ then the values change to $dy$ and $dx$. The slope is :
                $$lim_{triangle xrightarrow 0}frac{triangle y}{triangle x}=frac{dy}{dx}=m$$The visualization is like you bring the coordinate $B$ closer and closer to $A$ and keep on calculating slope until the point $B$ is so close that it coincides with the point $A$. Now you get the slope of the line joining $A$ and $A$ or in other words slope of the line at point $A$.



                Same things can be generalized for a function $f(x)$
                enter image description here
                Consider a function $f(x)$ and a point $A(x,y)$ on it. We take another point $B(x+triangle x,y+triangle y)$. Now we calculate the slope again of the line $AB$ as:
                $$frac{f(x+triangle x)-f(x)}{x+triangle x -x}=frac{f(x+triangle x)-f(x)}{triangle x}$$



                Now, the line $AB$ is actually a secant to the curve $f(x)$. Just bring the point B closer and closer to A that is keep decreasing $triangle x$ so that it becomes very small. Finally as $triangle x rightarrow 0$, the slope changes to $frac{dy}{dx}$ and the line $AB$ which was a secant back then has turned into a line $AA$ which means that it is a tangent at point $A$. That is why the slope $frac{dy}{dx}$ gives the slope of the tangent at any point.
                Finally,
                $$lim_{triangle x rightarrow 0 }frac{f(x+triangle)-f(x)}{triangle x}=frac{dy}{dx}$$



                This is also known as the first principle or Ab Initio method to compute the derivative.



                What else can you do once you know the slope of the line using differentiation. Anything and literally everything. Knowing the slopes of line will tell you the change in the slope is positive or negative giving the idea whether the function itself is increasing or decreasing function. Also if the slope of a tangent becomes $0$ at a point then there is a possibility of the maxima or minima at the same point. You can find the rate of change of quantity with respect to another like velocity or acceleration. The underlying concept is only one, the slope of the tangent.
                Hope this helps.....






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  Differentiation of any function gives you the slope of the tangent at that point. Why? Consider a line with the equation:
                  $$y=mx+c$$
                  Consider two points $A(x,y)$ and another point that is slightly far from A, $B(x+triangle x,y + triangle y)$.
                  ![enter image description here
                  where $triangle y$ is the change in the function value for a change corresponding to $triangle x$. In other words, the change in $y$ occurs due to a change in $x$.
                  $$y=mx+c$$
                  $$triangle y=triangle {mx +c}=triangle {(mx)} + triangle c= mtriangle x$$
                  So, now we calculate the slope of the line $AB$, the coordinates are $A(x,y)$ and $B(x+triangle x,y+triangle y)$. The slope is :
                  $$frac{triangle y}{triangle x}=frac{mtriangle x}{triangle x}=m$$
                  If you make the changes $triangle x$ and $triangle y$, very small tending to $0$ then the values change to $dy$ and $dx$. The slope is :
                  $$lim_{triangle xrightarrow 0}frac{triangle y}{triangle x}=frac{dy}{dx}=m$$The visualization is like you bring the coordinate $B$ closer and closer to $A$ and keep on calculating slope until the point $B$ is so close that it coincides with the point $A$. Now you get the slope of the line joining $A$ and $A$ or in other words slope of the line at point $A$.



                  Same things can be generalized for a function $f(x)$
                  enter image description here
                  Consider a function $f(x)$ and a point $A(x,y)$ on it. We take another point $B(x+triangle x,y+triangle y)$. Now we calculate the slope again of the line $AB$ as:
                  $$frac{f(x+triangle x)-f(x)}{x+triangle x -x}=frac{f(x+triangle x)-f(x)}{triangle x}$$



                  Now, the line $AB$ is actually a secant to the curve $f(x)$. Just bring the point B closer and closer to A that is keep decreasing $triangle x$ so that it becomes very small. Finally as $triangle x rightarrow 0$, the slope changes to $frac{dy}{dx}$ and the line $AB$ which was a secant back then has turned into a line $AA$ which means that it is a tangent at point $A$. That is why the slope $frac{dy}{dx}$ gives the slope of the tangent at any point.
                  Finally,
                  $$lim_{triangle x rightarrow 0 }frac{f(x+triangle)-f(x)}{triangle x}=frac{dy}{dx}$$



                  This is also known as the first principle or Ab Initio method to compute the derivative.



                  What else can you do once you know the slope of the line using differentiation. Anything and literally everything. Knowing the slopes of line will tell you the change in the slope is positive or negative giving the idea whether the function itself is increasing or decreasing function. Also if the slope of a tangent becomes $0$ at a point then there is a possibility of the maxima or minima at the same point. You can find the rate of change of quantity with respect to another like velocity or acceleration. The underlying concept is only one, the slope of the tangent.
                  Hope this helps.....






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Differentiation of any function gives you the slope of the tangent at that point. Why? Consider a line with the equation:
                    $$y=mx+c$$
                    Consider two points $A(x,y)$ and another point that is slightly far from A, $B(x+triangle x,y + triangle y)$.
                    ![enter image description here
                    where $triangle y$ is the change in the function value for a change corresponding to $triangle x$. In other words, the change in $y$ occurs due to a change in $x$.
                    $$y=mx+c$$
                    $$triangle y=triangle {mx +c}=triangle {(mx)} + triangle c= mtriangle x$$
                    So, now we calculate the slope of the line $AB$, the coordinates are $A(x,y)$ and $B(x+triangle x,y+triangle y)$. The slope is :
                    $$frac{triangle y}{triangle x}=frac{mtriangle x}{triangle x}=m$$
                    If you make the changes $triangle x$ and $triangle y$, very small tending to $0$ then the values change to $dy$ and $dx$. The slope is :
                    $$lim_{triangle xrightarrow 0}frac{triangle y}{triangle x}=frac{dy}{dx}=m$$The visualization is like you bring the coordinate $B$ closer and closer to $A$ and keep on calculating slope until the point $B$ is so close that it coincides with the point $A$. Now you get the slope of the line joining $A$ and $A$ or in other words slope of the line at point $A$.



                    Same things can be generalized for a function $f(x)$
                    enter image description here
                    Consider a function $f(x)$ and a point $A(x,y)$ on it. We take another point $B(x+triangle x,y+triangle y)$. Now we calculate the slope again of the line $AB$ as:
                    $$frac{f(x+triangle x)-f(x)}{x+triangle x -x}=frac{f(x+triangle x)-f(x)}{triangle x}$$



                    Now, the line $AB$ is actually a secant to the curve $f(x)$. Just bring the point B closer and closer to A that is keep decreasing $triangle x$ so that it becomes very small. Finally as $triangle x rightarrow 0$, the slope changes to $frac{dy}{dx}$ and the line $AB$ which was a secant back then has turned into a line $AA$ which means that it is a tangent at point $A$. That is why the slope $frac{dy}{dx}$ gives the slope of the tangent at any point.
                    Finally,
                    $$lim_{triangle x rightarrow 0 }frac{f(x+triangle)-f(x)}{triangle x}=frac{dy}{dx}$$



                    This is also known as the first principle or Ab Initio method to compute the derivative.



                    What else can you do once you know the slope of the line using differentiation. Anything and literally everything. Knowing the slopes of line will tell you the change in the slope is positive or negative giving the idea whether the function itself is increasing or decreasing function. Also if the slope of a tangent becomes $0$ at a point then there is a possibility of the maxima or minima at the same point. You can find the rate of change of quantity with respect to another like velocity or acceleration. The underlying concept is only one, the slope of the tangent.
                    Hope this helps.....






                    share|cite|improve this answer









                    $endgroup$



                    Differentiation of any function gives you the slope of the tangent at that point. Why? Consider a line with the equation:
                    $$y=mx+c$$
                    Consider two points $A(x,y)$ and another point that is slightly far from A, $B(x+triangle x,y + triangle y)$.
                    ![enter image description here
                    where $triangle y$ is the change in the function value for a change corresponding to $triangle x$. In other words, the change in $y$ occurs due to a change in $x$.
                    $$y=mx+c$$
                    $$triangle y=triangle {mx +c}=triangle {(mx)} + triangle c= mtriangle x$$
                    So, now we calculate the slope of the line $AB$, the coordinates are $A(x,y)$ and $B(x+triangle x,y+triangle y)$. The slope is :
                    $$frac{triangle y}{triangle x}=frac{mtriangle x}{triangle x}=m$$
                    If you make the changes $triangle x$ and $triangle y$, very small tending to $0$ then the values change to $dy$ and $dx$. The slope is :
                    $$lim_{triangle xrightarrow 0}frac{triangle y}{triangle x}=frac{dy}{dx}=m$$The visualization is like you bring the coordinate $B$ closer and closer to $A$ and keep on calculating slope until the point $B$ is so close that it coincides with the point $A$. Now you get the slope of the line joining $A$ and $A$ or in other words slope of the line at point $A$.



                    Same things can be generalized for a function $f(x)$
                    enter image description here
                    Consider a function $f(x)$ and a point $A(x,y)$ on it. We take another point $B(x+triangle x,y+triangle y)$. Now we calculate the slope again of the line $AB$ as:
                    $$frac{f(x+triangle x)-f(x)}{x+triangle x -x}=frac{f(x+triangle x)-f(x)}{triangle x}$$



                    Now, the line $AB$ is actually a secant to the curve $f(x)$. Just bring the point B closer and closer to A that is keep decreasing $triangle x$ so that it becomes very small. Finally as $triangle x rightarrow 0$, the slope changes to $frac{dy}{dx}$ and the line $AB$ which was a secant back then has turned into a line $AA$ which means that it is a tangent at point $A$. That is why the slope $frac{dy}{dx}$ gives the slope of the tangent at any point.
                    Finally,
                    $$lim_{triangle x rightarrow 0 }frac{f(x+triangle)-f(x)}{triangle x}=frac{dy}{dx}$$



                    This is also known as the first principle or Ab Initio method to compute the derivative.



                    What else can you do once you know the slope of the line using differentiation. Anything and literally everything. Knowing the slopes of line will tell you the change in the slope is positive or negative giving the idea whether the function itself is increasing or decreasing function. Also if the slope of a tangent becomes $0$ at a point then there is a possibility of the maxima or minima at the same point. You can find the rate of change of quantity with respect to another like velocity or acceleration. The underlying concept is only one, the slope of the tangent.
                    Hope this helps.....







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 52 mins ago









                    SNEHIL SANYALSNEHIL SANYAL

                    638110




                    638110























                        1












                        $begingroup$

                        Derivatives are important in modeling physical phenomena via differential equations. These equations can model real-world processes much more accurately than a purely algebraic one. For example, the equation $theta’’+frac g l sintheta=0$ models the motion of a simple pendulum, where $theta$ is the angular displacement, $l$ is the length of the pendulum, and $g$ is the acceleration due to gravity.






                        share|cite









                        $endgroup$


















                          1












                          $begingroup$

                          Derivatives are important in modeling physical phenomena via differential equations. These equations can model real-world processes much more accurately than a purely algebraic one. For example, the equation $theta’’+frac g l sintheta=0$ models the motion of a simple pendulum, where $theta$ is the angular displacement, $l$ is the length of the pendulum, and $g$ is the acceleration due to gravity.






                          share|cite









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Derivatives are important in modeling physical phenomena via differential equations. These equations can model real-world processes much more accurately than a purely algebraic one. For example, the equation $theta’’+frac g l sintheta=0$ models the motion of a simple pendulum, where $theta$ is the angular displacement, $l$ is the length of the pendulum, and $g$ is the acceleration due to gravity.






                            share|cite









                            $endgroup$



                            Derivatives are important in modeling physical phenomena via differential equations. These equations can model real-world processes much more accurately than a purely algebraic one. For example, the equation $theta’’+frac g l sintheta=0$ models the motion of a simple pendulum, where $theta$ is the angular displacement, $l$ is the length of the pendulum, and $g$ is the acceleration due to gravity.







                            share|cite












                            share|cite



                            share|cite










                            answered 6 mins ago









                            csch2csch2

                            4281312




                            4281312






























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