Easy Way to Understand the Chain Rule

July 30, 2022 Post a Comment

The chain rule in math is an essential derivative rule that enables us to manage composite functions. With this article we will aim to learn about:

What is the chain rule in differentiation?
How to apply the chain rule for the calculation?
Chain rule for derivatives and more with solved examples.

The chain rule says that the instantaneous rate of change of a function "f" that is relative to "g" which is relative to "x" helps us calculate the instantaneous rate of change of "f" with respect to "x".

Basically, the chain rule is applied to determine the derivatives of composite functions like\((x^2+2)^4,(\sin4x),(\ln7x),e^{2x}\), and so on. If a function is represented as\( y = f(g(x))\), then by chain rule derivative we get \(y^{\prime}=f^{\prime}(g(x)).g^{\prime}(x).\)

Let us read more about the chain rule formula and the steps followed in obtaining the derivatives applying the chain rule. Beginning with some basic definitions of function and derivation.

Also, learn about Relations and Functions here.

What is Chain Rule?

This chain rule is also recognised as; an outside-inside rule/the composite function rule/function of a function rule. It is used solely to find the derivatives of the composite functions.

The chain rule in calculus as read in the introduction is the primary method for differentiating a composite function. Consider if f(x) and g(x) are two functions, the composite function f(g(x)) is calculated for the value of x by first evaluating g(x) and followed by evaluating the function f at this value of g(x), and finally "chaining" the results together.

\(\frac{dy}{dx}=g\left(x\right),\ where\ y=f\left(x\right)\)

For example, if \(f(x)=\sin x\) and \(g(x)=x^{3}\), then \(f(g(x))=\sin x^3\), while\(g(f(x))=(\sin x)^3\).

For this example\(\sin (x)^3\), the rule gives the result as:\(D\left(\sin(x)^3\right)=D\sin(x)^3.\ D(x)^3=(\cos x)^3.3x^2\).

The chain rule in terms of differentiation is represented as :\(\frac{d(f(g(x)))}{dx}=\frac{df}{dg}.\frac{dg}{dx}\)

If g(x) is differentiable at a given point x and f(x) is differentiable at the point g(x), then fog is differentiable at x. Furthermore this can be understood as;

Let y=f(g(x)) and u=g(x), then:

\(\frac{dy}{dx}=\frac{dy}{du}.\frac{du}{dx}\).

The above chain rule formula is obtained using Leibniz's Notation.

Also, read about x-axis and y-axis here.

Chain Rule Steps

Step 1: Recognize the chain rule: The function needs to be a composite function, which implies one function is nested over the other one.

Step 2: Know the inner function and the outer function respectively.

Step 3: Determine the derivative of the outer function, dropping the inner function.

Step 4: Obtain the derivative of the inner function.

Step 5: Multiply the outcomes from step 4 and step 5 respectively.

Step 6: Simplify the obtained chain rule derivative.

For example:

Consider a function: \(g(x)=\ln(\cos x)\).

Here "g" is a composite function therefore we can apply the chain rule.
Next is cos x is the inner function and ln(x) denotes the outer function.
The derivative of the outer function is equivalent to\(\frac{1}{\cos x}\).
The derivative of the inner function is -sin x.
Lastly \(g^{\prime}(x)\)= derivative of the outside function, leaving the inside alone × the derivative of the inside function = \(\frac{1}{\cos x}\times\left(-\sin x\right)\).
On simplifying we obtain, \(\frac{-\sin x}{\cos x}=-\tan x\).

Also learn the various Applications of Derivatives here.

Chain Rule Formula and Proof

There are two forms of chain rule formula as discussed below.

Chain Rule Formula 1

\(\frac{d}{dx}(f(g(x))=f^{\prime}(g(x))·g^{\prime}(x)\).

Example 1: To find the derivative of \(\frac{d}{dx}(\sin4x)\), write sin 4x = f(g(x)), where f(x) = sin x and g(x) = 4x.

Then by applying the chain rule formula:

\(\frac{d}{dx}(\sin4x)=\cos4x·4=4\cos4x\)

Chain Rule Formula 2

We can assume the expression that is replacing "x" with "u" and applying the chain rule formula.

\(\frac{dy}{dx}=\frac{dy}{du}.\frac{du}{dx}\).

Example 2: To find \(\frac{d}{dx}(\sin4x)\), assume that y = sin 4x and 4x = u. Then y = sin u.

By the chain rule formula we obtain:

\(\frac{d}{dx}(\sin4x)=\frac{d}{du}(\sin u)·\frac{d}{dx}(4x)=\cos u·4=4\cos u=4\cos4x\)

Learn more about Integral Calculus here.

Double Chain Rule

There can be nested functions one inside the other or one over the other, where the functions rely on more than one variable. The chain of smaller derivatives is multiplied collectively to receive the overall derivative.

Consider there are three functions: p, q, r. A function f is a composite of p, q, and r. The chain rule is extended hereabouts. If a function is a combination of three functions, we use the chain rule twice.

That is when f = (p o q) o r = \(\frac{df}{dx}=\frac{df}{dp}.\ \frac{dp}{dq}.\ \frac{dq}{dr}.\ \frac{dr}{dx}\)

Example 3: y = \((1+\cos3x)^{3}\)

\(y^{\prime}=3(1+\cos3x)\times(-\sin3x)\times(3)\)

\(y^{\prime}=-9(1+\cos3x)\times(\sin3x)\)

So far what we have discussed is called the chain rule on univariate functions. Let us also learn about how to apply the chain rule to multivariate functions.

Learn more about Logarithmic functions here.

Example 4: Assume that we are now offered by a multivariate function of two independent variables, p and q, with each of these variables being dependent on different two independent variables, x and y:

h = g(p, q) = \(p^{2} + q^{3}\)

Where the functions, p = xy, and q = 2x – y.

Performing the chain rule here needs the computation of partial derivatives, as we are operating with multiple independent variables. Moreover, p and q will also act as our standard variables. The formulae that we will be operating with, defined for each input, are as follows:

\(\frac{dh}{dx}=\frac{dh}{dp}.\frac{dp}{dx}+\frac{dh}{dq}.\frac{dq}{dx}\)

\(\frac{dh}{dy}=\frac{dh}{dp}.\frac{dp}{dy}+\frac{dh}{dq}.\frac{dq}{dy}\)

Obtaining the values for the above formula:

\(\frac{dh}{dx}=\frac{dh}{dp}.\frac{dp}{dx}+\frac{dh}{dq}.\frac{dq}{dx}\)

\(\frac{dh}{dp}=2p\)

\(\frac{dp}{dx}=y\)

\(\frac{dh}{dp}.\frac{dp}{dx}=2p.y\)

\(\frac{dh}{dq}=3q^2\)

\(\frac{dq}{dx}=2\)

\(\frac{dh}{dq}.\frac{dq}{dx}=6q^2\)

\(\frac{dh}{dx}=\frac{dh}{dp}.\frac{dp}{dx}+\frac{dh}{dq}.\frac{dq}{dx}\)

\(\frac{dh}{dx}=\frac{dh}{dp}.\frac{dp}{dx}+\frac{dh}{dq}.\frac{dq}{dx}=2py+6q^2\)

Similarly:

\(\frac{dh}{dy}=\frac{dh}{dp}.\frac{dp}{dy}+\frac{dh}{dq}.\frac{dq}{dy}\)

\(\frac{dh}{dp}=2p\)

\(\frac{dp}{dy}=x\)

\(\frac{dh}{dp}.\frac{dp}{dx}=2p.x\)

\(\frac{dh}{dq}=3q^2\)

\(\frac{dq}{dy}=-1\)

\(\frac{dh}{dq}.\frac{dq}{dx}=-3q^2\)

\(\frac{dh}{dy}=\frac{dh}{dp}.\frac{dp}{dy}+\frac{dh}{dq}.\frac{dq}{dy}\)

\(\frac{dh}{dy}=\frac{dh}{dp}.\frac{dp}{dx}+\frac{dh}{dq}.\frac{dq}{dx}=2p.x-3q^2\)

Learn the concepts of Three Dimensional Geometry here.

The Chain and Power Rules Combined

For a function like \(h\left(x\right)=\left(g\left(x\right)\right)^n\) we need to combine the chain rule with the power rule:

For example if \( f\left(x\right)=x^n\) then by power rule we get \( f^{^{\prime}}\left(x\right)=nx^{n-1}\).

Therefore \(h\left(x\right)=\left(g\left(x\right)\right)^n\) can be represented as: \(h^{^{\prime}}\left(x\right)=n\left(g\left(x\right)\right)^{n-1}.g^{^{\prime}}\left(x\right)\)

Learn the concepts of Complex Numbers here.

Chain Rule: Common Mistake

Normally, the only method to differentiate a composite function is by applying the chain rule. If we do not understand that a function is composite and that the chain rule should be applied, we will not be able to differentiate precisely.

Similarly, applying the chain rule on a function that is not composite will also result in an incorrect derivative.
Even when a student notices that a function is composite, they might arrange the inner and the outer functions incorrectly. This will end in the wrong derivative.
For example, in the composite function \(\cos(x^{2})\) the outer function is cos(x), whereas the inner function is \(x^{2}\). Students are mostly confused in deciding the outer and inner function.

Another common mistake done by students is to only differentiate the outer function and leave the inner function, which results in an incorrect answer.

For the relation: \(\frac{d}{dx}(f(g(x))=f^{\prime}(g(x))·g^{\prime}(x)\).

If the students differentiate only \(f^{\prime}(g(x))\) and leave the second part. While the correct form is \(f^{\prime}(g(x))·g^{\prime}(x)\).

Check more topics of Mathematics here.

Applications of the Chain Rule

This chain rule has extensive applications in the areas of physics, chemistry, and engineering. Some of them are as follows:

To determine the time rate of change of the pressure.
To estimate the rate of change of distance between two moving targets/objects.
To obtain the state of an object that is moving to the right and left in a precise interval.
To conclude if a function is progressing or decreasing.
To observe the rate of change of the average molecular speed.
The method of applying the chain rule to univariate functions can be continued to multivariate ones respectively.
The application of the chain rule obeys a similar process, no matter how complicated the function is: get the derivative of the outer function first and then proceed inwards.
Besides the process, the application of other derivative rules or formulas might be needed.
Implementing the chain rule to multivariate functions needs the knowledge of partial derivatives.

Chain Rule Solved Examples

With the knowledge of chain rule definition in math, along with the various formulas, and application we are all set to practice some more solved for a better understanding of the topic:

Question 1: Differentiate \(cos(sin(x^2))\)with respect to x ?
Solution:
Applying the concept of chain rule:

Let y = f(v) be a differentiable function of v and v = g(x) be a differentiable function of x then \(\frac{dy}{dx}=\frac{dy}{dv}⋅\ \frac{dv}{dx}\)
Calculation:
Let y =\(\cos(\sin(x^2))\)
⇒ \(\frac{dy}{dx}=\frac{d}{dx}\cos(\sin(x^2))\)
⇒\(\frac{dy}{dx}=−\sin(\sin(x^2))\ \frac{d}{dx}(\sin(x^2))\)
⇒ \(\frac{dy}{dx}=−\sin(\sin(x^2))\cos(x^2)\ \frac{d}{dx}(x^2)\)
⇒ \(\frac{dy}{dx}=−\sin(\sin(x^2))\cos(x^2)2x\)
⇒ \(\frac{dy}{dx}=−2x\sin(\sin(x^2))\cos(x^2)\)

Question 2: Differentiate \(f(x)=2\sin^2x^2\cos x^2\) and find the value of f'(x).
Solution:
Concept:
Chain Rule (Differentiation by substitution): If y is a function of u and u is a function of x
\(\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}\)
Calculation:
Let \( u = x^{2}\)
\(\frac{du}{dx}=2x\)

\(f(x)=2\sin^2x^2\cos x^2\) can be written as:
\(g(u)=2\sin^2u\cos u\)

⇒ \(\frac{dg(u)}{du}=2\left[\cos u\ \frac{d\sin^2u}{du}+\sin^2u\frac{d\cos u}{du}\right]\)

⇒\(\frac{dg(u)}{du}=2\left[\left(2\sin u\cos u\right)\cos u+\sin^2u\left(-\sin u\right)\right]\)

⇒\(\frac{dg(u)}{du}=4\sin u\cos^2u-2\sin^{3} u\)
Now,

\(f^{\prime}(x)=\frac{df(x)}{dx}=\frac{dg(u)}{du}\times\frac{du}{dx}\)

\(f^{\prime}(x)=\left(4\sin u\cos^2u-2\sin^3u\right)\times2x\)
Now, substitute the value of \(u = x^{2}\) in f'(x)

\(f^{\prime}(x)=4x\left(2\sin x^2\cos^2x^2-\sin^3x^2\right)\)

Question 3: If \(y=e^{x+e^{x+e^{x+^{……∞}}}}\), then \(\frac{dy}{dx}\) is:
Solution:
Concept:
Chain Rule of Derivatives: .
\(\frac{d}{dx}f(g(x))=\frac{d}{dg(x)}f(g(x))\times\frac{d}{dx}g(x)\)
\(\frac{d}{dx}e^x=e^x\).
Calculation:
It is given that \(y=e^{x+e^{x+e^{x+^{……∞}}}}\).
∴ \(y=e^{x+\left(e^{x+e^{x+^{……∞}}}\right)}=e^{x+y}\)
Differentiating both sides with respect to x and using the chain rule, we get:
\(\frac{dy}{dx}=\frac{d}{dx}e^{x+y}\)
⇒
\(\frac{dy}{dx}=e^{x+y}\frac{d}{dx}\left(x+y\right)\)
⇒ \(\frac{dy}{dx}=y\left(1+\frac{dy}{dx}\right)\)
⇒ \(\frac{dy}{dx}=y+y\frac{dy}{dx}\)
⇒ \(\left(1-y\right)\frac{dy}{dx}=y\)
⇒\(\frac{dy}{dx}=\frac{y}{\left(1-y\right)}\)

We hope that the above article on Chain Rule is helpful for your understanding and exam preparations. Stay tuned to the Testbook App for more updates on related topics from Mathematics, and various such subjects. Also, reach out to the test series available to examine your knowledge regarding several exams.

Chain Rule FAQs

Q.1 What is the chain rule?

Ans.1 In mathematics, the chain rule is defined as a formula for calculating the derivative of the composition of two or more functions.

Q.2 When can you use the chain rule?

Ans.2 We practice the chain rule when differentiating a 'function of a function, for example, f(g(x)) in general.

Q.3 Why is it called the chain rule?

Ans.3 This rule is identified as the chain rule because we utilise it to take derivatives of composites of functions by connecting them together with their derivatives. The chain rule can be considered as taking the derivative of the outer function connected to the inner function and then multiplying it by the derivative of the inner function.

Q.4 What are some of the applications of derivatives in maths?

Ans.4 In maths, derivatives have extensive applications. They are applied in many circumstances like calculating the slope of the curve, determining the maxima or minima of a function, obtaining the slope of the curve, and also inflection points.

Q.5 What is the definition of derivatives?

Ans.5 The derivatives of the function determine the rate of change of a function at a point, and therefore also defined as the ratio of the change in the value of a dependent variable to the change in the independent variable which is expressed by dy/dx.

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