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?
$begingroup$
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
calculus soft-question education
$endgroup$
|
show 2 more comments
$begingroup$
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
calculus soft-question education
$endgroup$
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
|
show 2 more comments
$begingroup$
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
calculus soft-question education
$endgroup$
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
calculus soft-question education
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
|
show 2 more comments
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
|
show 2 more comments
4 Answers
4
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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, $B(x+triangle x,y + triangle y)$.
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 $f(x)$
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.....
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered 2 hours ago
marty cohenmarty cohen
74.3k549128
74.3k549128
add a comment |
add a comment |
$begingroup$
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, $B(x+triangle x,y + triangle y)$.
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 $f(x)$
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.....
$endgroup$
add a comment |
$begingroup$
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, $B(x+triangle x,y + triangle y)$.
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 $f(x)$
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.....
$endgroup$
add a comment |
$begingroup$
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, $B(x+triangle x,y + triangle y)$.
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 $f(x)$
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.....
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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, $B(x+triangle x,y + triangle y)$.
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 $f(x)$
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.....
answered 52 mins ago
SNEHIL SANYALSNEHIL SANYAL
638110
638110
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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.
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add a comment |
$begingroup$
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.
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add a comment |
$begingroup$
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.
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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.
answered 6 mins ago
csch2csch2
4281312
4281312
add a comment |
add a comment |
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3
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Possible duplicate of Applications of derivatives outside mathematics and physics
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– Don Thousand
2 hours ago
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@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...
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– rash
2 hours ago
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Sure, but that question definitely answers this.
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– Don Thousand
2 hours ago
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The answer to this kind of question is dynamic. Who knows maybe I will a different variety of answers @DonThousand
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– rash
2 hours ago
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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.
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– littleO
1 hour ago