Differentiation Using Chain Rule or Substitution: Definition, Formula, Reverse Chain

Differentiation has its uses in Mathematics, such as finding the rate of change of a quantity, finding the approximation value, finding the equation of tangent and normal to a curve, and finding the minimum and maximum values of algebraic expressions. Derivatives are vastly used across fields like Science, Engineering, and Physics.

While learning about derivatives, we have come across various methods, and among them is differentiation using the chain rule for composite functions. In this article, we will learn everything about differentiation using chain rule or substitution.

Chain Rule: Formula 

Let \(h\) be a real-valued composite function of two functions \(f\) and \(g\), i.e. \(h= f o g.\) Suppose \(u=g(x),\) where \(\frac{{du}}{{dx}}\) and \(\frac{{df}}{{du}}\) exist; then this could be expressed as

\(\left( {\frac{{{\rm{ change}}\,{\rm{in }}\,h}}{{{\rm{ change}}\,{\rm{in }}\,x}}} \right) = \left( {\frac{{{\rm{ change}}\,{\rm{in }}\,f}}{{{\rm{ change}}\,{\rm{in }}\,u}}} \right) \times \left( {\frac{{{\rm{ change}}\,{\rm{in }}\,u}}{{{\rm{ change}}\,{\rm{in }}\,x}}} \right)\)

\( \Rightarrow \frac{{dh}}{{dx}} = \frac{{df}}{{du}} \cdot \frac{{du}}{{dx}}\)

We often come across the functions where \(y\) is a function of \(u,\) and \(u,\) is a function of \(x.\) If \(y=f(u)\) and \(u=∅(x),\) then \(y=f[∅(x)].\) Such a function \(y\) is called a function of a function.

Differentiating this function of function, we get 

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

More generally, if \(y = f(u),u = \emptyset (v),\) and \(v = \varphi (x),\) then
\(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dv}} \cdot \frac{{dv}}{{dx}}\)

This is called the chain rule.

Example: Let’s differentiate sin \({x^4}\) w.r.t. \(x.\) 

Here, first we differentiate the trigonometric function, i.e. sin keeping \({x^4}\) as an independent variable, and then differentiate \({x^4}\) w.r.t \(x.\)

Step 1: Let’s use the chain rule to differentiate \(\sin {x^4}\) w.r.t. \(x.\)

Let \(y = \sin {x^4}\)

If we assume \(u = {x^4},\) here \(u\) is a function of \(x,\) and \(y\) becomes the function of the \(u.\)

Then \(y = \sin u.\)

Therefore, \(\frac{{dy}}{{du}} = \frac{d}{{du}}(\sin u) = \cos u\)             \(\left[ \because {\frac{d}{{dx}}(\sin x) = \cos x} \right]\)

Step 2: \(\frac{{du}}{{dx}} = \frac{d}{{dx}}\left( {{x^4}} \right) = 4{x^3}\)                           \(\left[ \because {\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n – 1}}} \right]\)

\(\therefore \frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}}\)

Step 3: Put the value of \(\frac{{dy}}{{du}}\) and \(\frac{{du}}{{dx}}\) in the above equation, we get

\(\frac{{dy}}{{dx}} = \cos u \cdot 4{x^3}\) 

Step 4: Substituting the value of \(u\) in the above equation, we get

\(\frac{{dy}}{{dx}} = \cos {x^4} \cdot 4{x^3}\)

Hence, the differentiation of \(\sin {x^4}\) w.r.t. \(x\) is \(\cos {x^4} \cdot 4{x^3}\)

Differentiation of Composite Functions

The composition of two functions, \(f(x)\) and \(g(x),\) where \(f(x)\) is acting as the first function and \(g(x)\) as the second function, can be represented by \(f(g(x))\) or \((f ∘ g)(x).\) It is a combination of two or more functions to get the result in another function. We can say that the composite functions are those functions that are written in terms of another function. 

To get the derivative of the composite function, differentiate the first function with respect to the second function, and then differentiate the second function with respect to its variable. 

To differentiate \(f(g(x)),\) first, differentiate \(f\) w.r.t. \(g(x),\) and then differentiate \(g(x)\) w.r.t \(x.\) This can be represented as,

\(\frac{d}{{dx}}[f(g(x))] = {f^\prime }(g(x)) \cdot {g^\prime }(x)\)

Proof: Let \(u = g(x);y = f(u)\)

Then \(\Delta u = g(x + \Delta x) – g(x)\)

And \(\Delta y = f(u + \Delta u) – f(u)\)

Therefore, \(\frac{{\Delta y}}{{\Delta x}} = \frac{{\Delta y}}{{\Delta u}} \times \frac{{\Delta u}}{{\Delta x}} = \frac{{f(u + \Delta u) – f(u)}}{{\Delta u}} \times \frac{{g(x + \Delta x) – g(x)}}{{\Delta x}}\)

As \(\Delta x \to 0,\Delta u \to 0\)

Thus  \(\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta u}} \times \frac{{\Delta u}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{f(u + \Delta u) – f(u)}}{{\Delta u}} \times \mathop {\lim }\limits_{\Delta x \to 0} \frac{{g(x + \Delta x) – g(x)}}{{\Delta x}}\)

\( = {f^\prime }(u) \times {u^\prime }(x)\)

\( = {f^\prime }(g(x)){g^\prime }(x)\)

Thus  \(\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = {f^\prime }(g(x)){g^\prime }(x)\)

Derivative of the composite function is nothing but the product of the derivative of the outer function with respect to the inner function (i.e., derivative of \(f\) w.r.t \(g\)) and the derivative of the inner function with respect to the variable.

Differentiation of Parametric Functions

Sometimes, \(x\) and \(y\) are given as functions of a single variable, i.e. \(x = \emptyset (t),y = \varphi (t),\) where \(x\) and \(y\) are two functions, and \(t\) is a variable. Here, \(x\) and \(y\) are called parametric functions or parametric equations, and \(t\) is called the parameter. 

To find \(\frac{{dy}}{{dx}}\) of parametric functions, we first obtain the relationship between \(x\) and \(y\) by eliminating the parameter \(t,\) and then, we differentiate it with respect to \(x.\) But, every time it may not be convenient to eliminate the parameter.

Therefore, \(\frac{{dy}}{{dx}}\) can also be obtained by the formula 

\(\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\)

Implicit Differentiation Using Chain Rule

The chain rule and implicit differentiation are used to differentiate complex equations.

We can use chain rule in explicit functions and implicit functions. 

Explicit Function: The function in which \(y\) can be directly written in terms of \(x\) or \(y=f(x),\) then \(y\) is called the explicit function. 

Example: \(y = x\cos x,\) and \(y = 3{x^2}\) 

Implicit Function: If the relation between the variables \(x\) and \(y\) are given by an equation containing both \(x\) and \(y,\) then this equation is not immediately solvable for \(y.\) The function \(y\) is called an implicit function of \(x,\) or we can say that implicit functions are those in which \(y\) cannot be expressed exclusively in terms of \(x.\)

We can write the implicit functions as \(f(x, y)=0,\) which is a function of \(x, y,\) expressed as an equation with the variable on one side.

Example: \({y^2} + xy = 0\) and \({x^2} + xy + y = 0\)

Note: Every explicit function \(y=f(x)\) can also be represented by an implicit function \(y-f(x)=0.\)

For differentiation of implicit functions \(f(x, y)=0\) or to obtain \(\frac{{dy}}{{dx}}\) we follow the following steps.

• Step 1: Consider \(y\) as \(y(x)\) to show that \(y\) is a function of \(x.\)
• Step 2: Differentiate the entire function with respect to \(x,\) as \(y\) is a function of \(x.\) This means that the derivative of every \(y\) is \(\frac{{dy}}{{dx}}.\)
• Step 3: Collect the coefficients of \(\frac{{dy}}{{dx}}\) on one side of the equation and transfer the remaining terms to the other side.
• Step 4: Simplify and find \(\frac{{dy}}{{dx}}\) in terms of \(x\) and \(y.\)

Note: 

While using the implicit differentiation, we have to keep in mind that it is applicable only if the function is differentiable.

Reverse Chain Rule: Formula

We know that integration and differentiation are reverse processes of each other. A beneficial application of the reverse rule formula is that derivative results can be stated as integration results by reversing the process. So, we can say that the reverse chain rule is a special method for integrating a function with two components, where one component is the derivative of the other.

If we have a function \(g(f(x)),\) where \(f\) is differentiable at \(x,\) and \(g\) is differentiable at \(f(x),\) then the differentiation of the function by using chain rule is 

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

Now, integrate both sides of this result with respect to ? by the reverse process of the differentiation of the function by the chain rule.

\(\int {\frac{d}{{dx}}} g(f(x))dx = \int g’ (f(x)) \cdot {f^\prime }(x)dx\)

\(g(f(x)) = \int {{g^\prime }} (f(x)) \cdot {f^\prime }(x)dx\)

\(\therefore \int {g’\left( {f\left( x \right)} \right) \cdot f’\left( x \right)dx = g\left( {f\left( x \right)} \right) + c} \)

Solved Examples

Q.1. Find \(\frac{{dy}}{{dx}}\) for the function \({x^6} + {y^6} – 6xy = 2\)
Ans: Here, \({x^6} + {y^6} – 6xy = 2,\) is not immediately solvable for \(y.\) The function is an implicit function.
To obtain \(\frac{{dy}}{{dx}}\) differentiate the entire function with respect to \(x,\) treating \(y\) as a function of \(x.\)
\(6{x^5} + 6{y^5}\frac{{dy}}{{dx}} – 6\left[ {x\frac{{dy}}{{dx}} + y \cdot 1} \right] = 0\)
\({x^5} + {y^5}\frac{{dy}}{{dx}} – \left[ {x\frac{{dy}}{{dx}} + y} \right] = 0\)
\({x^5} – y + \frac{{dy}}{{dx}}\left[ {{y^5} – x} \right] = 0\)
\(\frac{{dy}}{{dx}}\left[ {x – {y^5}} \right] = {x^5} – y\)
\(\therefore \frac{{dy}}{{dx}} = \frac{{{x^5} – y}}{{x – {y^5}}}\)

Q.2. Find \(\frac{{dy}}{{dx}}\) for the function \(y = \log (\cos x).\)
Ans: \(y = \log (\cos x)\) is a composition of the two functions.
Let \(g(x) = \cos x\) and let \(f(x) = \log (x)\) then \(f(g(x)) = \log (\cos x).\)
\(\frac{d}{{dx}}f(g(x)) = {f^\prime }(g(x)) \cdot {g^\prime }(x)\)
Differentiation of \(\log (\cos x)\) with respect to \(\cos x\) is \(\frac{1}{{\cos x}}.\)
Differentiation of \(\cos x\) with respect to \(x\) is \({ – \sin x}.\)
Then \(\frac{d}{{dx}}f(g(x)) = \frac{1}{{\cos x}} \cdot ( – \sin x).\)
\(\frac{d}{{dx}}\log (\cos x) = \frac{{ – \sin x}}{{\cos x}} = – \tan x\)
\(\therefore \frac{{dy}}{{dx}} = – \tan x.\)

Q.3. If \(x=2\) \(\sin t – \sin 2t\) and \(y = 2\cos t – \cos 2t,\) find the value of \(\frac{{dy}}{{dx}}.\)
Ans: \(x\) and \(y\) are given as functions of \(t.\)
Therefore, \(\frac{{dy}}{{dx}},\) can be obtained by the formula \(\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}.\)
\(\frac{{dx}}{{dt}} = 2\cos t – 2\cos 2t = 2(\cos t – \cos 2t)\)
And, \(\frac{{dy}}{{dt}} = – 2\sin t + 2\sin 2t = 2( – \sin t + \sin 2t)\)
\(\therefore \frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{{2( – \sin t + \sin 2t)}}{{2(\cos t – \cos 2t)}} = \frac{{( – \sin t + \sin 2t)}}{{(\cos t – \cos 2t)}}\)
\(\therefore \frac{{dy}}{{dx}} = \frac{{( – \sin t + \sin 2t)}}{{(\cos t – \cos 2t)}}\)

Q.4. If \(y = {v^3},v = 3u + 1\) and \(u = {x^2} + 2,\) then find \(\frac{{dy}}{{dx}}\) using chain rule.
Ans: Here, \(y\) is a function of \(v, v\) is a function of \(u,\) and \(u\) is a function of \(x.\)
So, differentiate \(y = {v^3}\) w.r.t. \(v,v = 3u + 1\) w.r.t. \(u\) and \(u = {x^2} + 2\) w.r.t \(x\), and by using chain rule.
\(\frac{{dy}}{{dx}} = \frac{{dy}}{{dv}} \cdot \frac{{dv}}{{du}} \cdot \frac{{du}}{{dx}}……..\left( i \right)\)
Step 1: \(y = {v^3}\)
Differentiate \(y\) w.r.t. \(v,\) we get,
\(\frac{{dy}}{{dv}} = 3{v^2}\,\,\,\,\,\left[ \because {\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n – 1}}} \right]\)
Step 2: \(v=3u+1\)
Differentiate \(v\) w.r.t. \(u,\) we get
\(\frac{{dv}}{{du}} = 3\,\,\,\,\,\,\,\,\left[ \because {\frac{d}{{dx}}\left( {k{x^n} + c} \right) = kn{x^{n – 1}}} \right.\) Where \(k\) and \(c\) are constants
Step 3: \(u = {x^2} + 2\)
Differentiate \(u\) w.r.t. \(x,\) we get
\(\frac{{du}}{{dx}} = 2x\quad \left[ \because {\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n – 1}},\frac{d}{{dx}}(} \right.{\rm{constant}}\left. {) = 0} \right]\)
Step 4: By putting the values from step 1, step 2, and step 3 in equation \((i),\) we get,
\(\frac{{dy}}{{dx}} = \frac{{dy}}{{dv}} \cdot \frac{{dv}}{{du}} \cdot \frac{{du}}{{dx}} = 3{v^2} \cdot 3 \cdot 2x\)
\(\therefore \frac{{dy}}{{dx}} = 3{v^2} \cdot 3 \cdot 2x\)
By substituting the value of \(v\) and \(u\) in the above equation, we get,
\(\frac{{dy}}{{dx}} = 3{\left( {3\left( {{x^2} + 2} \right) + 1} \right)^2} \cdot 3 \cdot 2x\)
\(\frac{{dy}}{{dx}} = 3{\left( {3{x^2} + 7} \right)^2} \cdot 3 \cdot 2x\)
\(\frac{{dy}}{{dx}} = 18x\left( {9{x^4} + 49 + 42{x^2}} \right)\)
\(\frac{{dy}}{{dx}} = 162{x^5} + 882x + 756{x^3}\)
\(\therefore \frac{{dy}}{{dx}} = 162{x^5} + 756{x^3} + 882x\)

Q.5. By using reverse chain rule formula find \(\int 4 {x^3}\cos {x^4}dx.\)
Ans: Let \(g(x) = \sin x\) and \(f(x) = {x^4}\)
Now, we have a function \(g(f(x)) = \sin {x^4}\)
Then the differentiation of the function by using chain rule is
\(\frac{d}{{dx}}g(f(x)) = {g^\prime }(f(x)) \cdot {f^\prime }(x) = 4{x^3}\cos {x^4}\)
Now, we will integrate both sides of this result with respect to ? by the reverse process of the differentiation of the function by chain rule
\(\int {\frac{d}{{dx}}} g(f(x))dx = \int {{g^\prime }} (f(x)) \cdot {f^\prime }(x)dx\)
\(g(f(x)) + c = \int {{g^\prime }} (f(x)) \cdot {f^\prime }(x)dx\)
Hence, \(\int 4 {x^3}\cos {x^4}dx = \sin {x^4} + c.\)

Summary

Let \(h\) be a real-valued composite function of two functions \(f\) and \(g,\) i.e. \(h= f o g.\) Suppose \(u=g(x),\) where \(\frac{{du}}{{dx}}\) and \(\frac{{df}}{{du}}\) exist, then this could be expressed as \(\frac{{dh}}{{dx}} = \frac{{df}}{{du}}.\frac{{du}}{{dx}}\) When \(x\) and \(y\) are parametric functions of variable \(t,\) then \(\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\) The function in which y can be directly written in terms of \(x\) or \(y=f(x),\) then \(y\) is called the explicit function. Implicit functions are those in which \(y\) cannot be expressed exclusively in terms of \(x \cdot \int {{g^\prime }} (f(x)) \cdot {f^\prime }(x)dx = g(f(x)) + c\) is the reverse process of the chain rule.

Frequently Asked Questions (FAQs)

Q.1. How do you use chain rule in differentiation?
Ans: If \(y=f(u)\) is the function of \(u\) and \(u=∅(x)\) is the function of \(x,\) then
\(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}}\)

Q.2. How do you use the substitution method in differentiation?
Ans: The substitution method is useful to reduce the functions into the simplest form. For problems involving inverse trigonometric functions, first, use a suitable substitution to simplify it and then differentiate it.

Q.3. How do you do the chain rule step by step?
Ans: The chain rule is done as follows:
Step 1: First identify the inner function and rewrite the outer function by replacing the inner function using any suitable variable.
Step 2: Find the derivative of both functions separately with respect to their variables.
Step 3: Now substitute the derivatives and the original expression for the assumed variable into the Chain Rule.
Step 4: Simplify it and write its simplest form.

Q.4. How do you use the chain rule for trigonometry functions?
Ans: We have come across to many questions involving trigonometry functions. To simplify it, make a suitable substitution in the form of another function and then use the chain rule for the differentiation.
Example: To find the derivative of \(\sin \left( {2{x^2} + 5x + 1} \right)\) make a substitution \(u = 2{x^2} + 5x + 1,\) then differentiate \(\sin \) w.r.t. \(u\) and \(u\) w.r.t. \(x\) and then substitute all together in the formula of chain rule.

Q.5. What is the difference between the chain rule and the power rule?
Ans: The power rule is nothing but a special case of the chain rule. It is useful to find the derivative of a function which raised to the \({n^{{\rm{th}}}}\) power. The power rule states that the derivative of a function is \(n\) times the function raised to the \({(n – 1)^{{\rm{th}}}}\) power times the derivative of the function.

We hope this detailed article on the Differentiation Using Chain Rule or Substitution will make you familiar with the topic. If you have any inquiries, feel to post them in the comment box. Stay tuned to embibe.com for more information.