Chain Rule of Differentiation: Steps and Solved Examples

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The Chain Rule formula computes the derivative of the composite of two or more functions. The chain rule for the differentiation of composite functions is defined. For example, if f and g are functions, then the chain rule expresses their composition's derivative.

d/dx [f(g(x))] = f'(g(x)) g'(x)

Key Terms: Derivative, Function, Constant, Trigonometric, Calculus, Limits, Continuity, Logarithmic, Exponential


What is Differentiation?

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Rates of change can be found by using differentiation. Differentiation lets us find, for example, the rate of change of velocity in relation to time (which gives the acceleration).

  • The idea of differentiation also lets us find the rate of change of the variable x in relation to the variable y. 
  • This rate is called the gradient of the curve when it is shown on a graph of y against x. 
  • The chain rule can be written as f o g, where f and g are two functions. 
  • Now to find the difference between functions that are known to be composite there are a few steps:
  1. For instance, let's look at the function (x + 3)2. The inner function, g, is equal to (x + 3), and if x + 3 = u, then the outer function, f, can be written as u2 times the inner function.
  2. This rule is also called the chain rule because it is used to find the derivatives of functions that are made up of other functions. This is done by chaining the derivatives of the functions together.
  3. The chain rule can be thought of as multiplying the derivative of the outer function (which is used to calculate the inner function) by the derivative of the inner function.

d/dx[f(x)n]=n(f(x))n−1.f’(x)

d/dx[f(x)n]=f’(g(x))g’(x)

Read More: Non-Parametric Test

Discover about the Chapter video:

Continuity and Differentiability Detailed Video Explanation:


What is Chain Rule Formula?

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When one multiplies the derivative of the outer function by the derivative of the inner function, we get the derivative of the composite function.

Here are the two functions f(x) and g(x), The chain rule formula is,

(fog )( x ) = f ′ ( g( x ) )·g′( x )

Read More: Infinity


Steps to Solve Chain Rule Formula

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Here are the terms f’(x) and g’(x) represent the differentiation of the functions f(x) and g(x), respectively.

  1. In order to produce f'(g(x)), one will need to differentiate the outer function while maintaining the consistency of the inner function.
  2. One must now differentiate the inner function in order to compute the function g'(x), which stands for "x."
  3. Now all that is required of you is to express the final answer you have gotten in a form that is simple.

Read More: Differentiation Rules


Common Mistakes Made in Chain Rule

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Here are some of the most common mistakes students make when using the chain rule formula.

  • One often misidentifies a function as composite since the chain rule is the sole way to separate a composite function. 
  • Without it, the function will be erroneously derived.
  • After realizing the function is composite, one may misidentify inner and outer functions, resulting in an incorrect derivative.
  • One commonly forgets to multiply by the derivative of the inner function while applying the chain rule. 
  • One usually differentiates the outer function without deriving the inner function, which is incorrect.

Solved Examples

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Example 1: Compute the derivative of the function y = arcsin(2x + 1), i.e., y = sin-1(2x + 1).

Solution: Given,

y = arcsin(2x + 1)

= sin-1(2x + 1)

Let y = sin-1(u) and u = 2x + 1.

dy/du = (d/du) sin-1(u)

= 1/√(1 – u2)

du/dx = (d/dx)(2x + 1)

= 2

Using the chain rule;

dy/dx = (dy/du)(du/dx)

= [1/√(1 – u2)] (2)

= 2/√[1 – (2x + 1)2]

= 2/√[1 – (4x2 + 1 + 4x)]

= 2/√(1 – 4x2 – 1 – 4x)

= 2/√( – 4x2 – 4x)

= 2/2√[-x(x + 1)]

= 1/√[-x(x + 1)]

Hence, dy/dx = 1/√[-x(x + 1)].

Example 2: Find dy/dx if y = 2 sin(3 cos 4x).

Solution: Given,

y = 2 sin(3 cos 4x)

Let u = 3 cos 4x and y = 2 sin u

du/dx = 3(-sin 4x)(d/dx)4x

= -12 sin 4x

And dy/du = (d/du) (2 sin u) = 2 cos u

Using the chain rule;

dy/dx = (dy/du) (du/dx)

= (2 cos u) (-12 sin 4x)

= -24 sin 4x cos(3 cos 4x)

Therefore, dy/dx = -24 sin 4x cos(3 cos 4x).

Read More: Differential Equations


Things to Remember

  • The chain rule differentiates composite functions. Consequently, h'(x) = f'(g (x)). g’(x). 
  • Leibniz's chain rule is dy/dx=(dy/du)*(du/dx).
  • Formulas can be derived using the chain rule and other principles.
  • Combining the chain and power rules creates a new rule. 
  • h'(x) is the product of g'(x) and n(g(x)) to the power of n. (n - 1).
  • In order to produce f'(g(x)), one will need to differentiate the outer function while maintaining the consistency of the inner function.
  • If the chain rule is applied to the composition of three functions, the rule h(x) = f(g(k(x))) becomes h'(x) = f'(g(k(x))). g’(k(x)) . k’(x)

Previous Year Questions

  1. Nascent hydrogen consists of..
  2. In which of the following reactions the hydrogen peroxide acts as a reducing agent?..[JEE MAIN 2023]
  3. Major product of the following reaction is..[JEE MAIN 2023]
  4. The total current supplied to the circuit by the battery is...[AIEEE 2004]
  5. The reading of voltmeter in the circuit shown is….[Rajasthan PMT 2023]
  6. The oxidation of toluene to benzaldehyde by chromyl chloride is called...[NEET UG 1996]
  7. An aggregate fruit is one which develops from..[NEET UG 2014]
  8. The order of stability of the following carbocations..[JEE MAIN 2013]
  9. Complete hydrolysis of cellulose gives​...[BITSAT 2012]
  10. Which one among the following metals is the weakest reducing agent?..[JEE MAIN 2023]

Sample Questions

Ques. Differentiate y = cos x2 (2 Marks)

Ans. y = cos x2

Let u = x2 , so that y = cos u

Therefore;

du/dx = 2x

dy/du = -sinu

And so, the chain rule says:

dy/dx = dy/du * du/dx

dy/dx = – sinu * 2x = -2x sinx2

Ques. Differentiate f(x) = (1 + x2)5. (2 Marks)

Ans. Using the Chain rule,

dy/dx = dy/du ⋅ du/dx

Let us take y = u5 and u = 1 + x2

Then dy/du = d/du (u5) = 5u4

du/dx = d/dx (1 + x2 )= 2x

dy/dx = 5u4⋅2x = 5(1 + x2)4⋅2x

= 10x(1 + x2)4

Ques. If y = cos x3, find dy/dx. (2 Marks)

Ans. Given,

y = cos x3

Let y = cos u and u = x3

dy/du = – sin u

du/dx = 3x2

Now, using the chain rule, we have;

dy/dx = (dy/du) (du/dx)

= (-sin u) (3x2)

= – 3x2 sin x3 {since u = x3)

Therefore, dy/dx = – 3x2 sin x3.

Ques. Differentiate the function y = 4 ln(√x) with respect to x. (3 Marks)

Ans. Given,

y = 4 ln(√x)

Let y = 4 ln(u) and u = √x

dy/du = (d/du) 4 ln(u) = 4(1/u) = 4/u

du/dx = (d/dx)(√x) = 1/2√x

Using the chain rule of differentiation, 

dy/dx = (dy/du) (du/dx)

= (4/u) (1/2√x)

= (4/2) (1/√x.√x)

= 2/x

Therefore, dy/dx = 2/x.

Ques. Find the derivative of the function y = tan(4x3 – 5x + 2). (3 Marks)

Ans. Given,

y = tan(4x3 – 5x + 2)

Let y = tan u and u = 4x3 – 5x + 2

dy/du = (d/du) tan u

= sec2u

du/dx = (d/dx) (4x3 – 5x + 2)

= 12x2 – 5

Thus, by chain rule of differentiation;

dy/dx = (dy/du) (du/dx)

= (sec2u) (12x2 – 5)

= (12x2 – 5) sec2(4x3 – 5x + 2)

Hence, the derivative of y = tan(4x3 – 5x + 2) is dy/dx = (12x2 – 5) sec2(4x3 – 5x + 2).

Ques. What is the first derivative of the function f(x) = [cos(7x2 + 3)]4 (3 Marks)

Ans. Given, f(x) = [cos(7x2 + 3)]4

Let f(x) = y = (cos u)4 and u = 7x2 + 3

Consider y = (cos u)4

Let y = t4 and t = cos u

dy/dt = (d/dt)t4 = 4t3

dt/du = (d/du) cos u = – sin u

Now, by applying the chain rule, we get;

dy/du = (4t3) ( – sin u)

= – 4(cos u)3 (sin u)

= – 4[cos(7x2 + 3)]3 [sin(7x2 + 3)]

Consider u = 7x2 + 3

du/dx =(d/dx)(7x2 + 3) = 14x

Therefore, f’(x) = dy/dx = (dy/du)(du/dx) {using the chain rule}

= – 4[cos(7x2 + 3)]3 [sin(7x2 + 3)] (14x)

= -56x [cos(7x2 + 3)]3 [sin(7x2 + 3)]

Ques. Find the derivative of y = √[(x + 3)3(x – 1)4]. (3 Marks)

Ans. Given,

y = √[(x + 3)3(x – 1)4]

Let y = √t and t = (x + 3)3(x – 1)4

dy/dt = (d/dt)√t = 1/(2√t)

dt/dx = (d/dx)uv, where u = (x + 3)3 and v = (x – 1)4

= u(dv/dx) + v(du/dx)

= 4(x + 3)3(x − 1)3 + 3(x − 1)4(x + 3)2

= (x + 3)2(x − 1)3(7x + 9)

Using the chain rule;

dy/dx = (dy/dt) (dt/dx)

= [1/(2√t)] [(x + 3)2(x − 1)3(7x + 9)]

= [(x + 3)2(x − 1)3(7x + 9)]/ 2√[(x + 3)3(x – 1)4]

= (½) (x – 1)(7x + 9) √(x + 3)

Hence, the derivative of y = √[(x + 3)3(x – 1)4] is dy/dx = (½) (x – 1)(7x + 9) √(x + 3).

Ques. Using the chain rule, find the derivative of the function y = 1/(x3 + 4x2 − 3x − 3)6(3 Marks)

Ans. Given function is

y = 1/(x3 + 4x2 − 3x − 3)6

Ley y = 1/t6 and t = x3 + 4x2 – 3x – 3.

dy/dy = (d/dt)1/t6= – 6/t7 = – 6t-7

dt/dx = (d/dx)(x+ 4x2 – 3x – 3)

= 3x2 + 8x – 3

Using the chain rule,

dy/dx = (-6t-7) (3x2 – 8x – 3)

= -6(3x2 – 8x – 3) (x3 + 4x2 − 3x − 3)-7

Thus, dy/dx = – 6(3x2 – 8x – 3) (x3 + 4x2 − 3x − 3)-7.

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CBSE CLASS XII Related Questions

  • 1.
    Find the vector and cartesian equations of the line passing through the point of intersection of the lines \( \vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j}) \) and \( \vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k}) \) and parallel to the line \( \frac{x - 1}{-2} = \frac{7 - y}{-3} = z \).


      • 2.
        Which of the following equations is NOT a Linear Differential Equation?

          • \((1 + x^2) \, dy + 2xy \, dx = \cot x \, dx\)
          • \(y + \frac{d}{dx}(xy) = x(\sin x + \log x)\)
          • \(x(1 + y^2) \, dx - y(1 + x^2) \, dy = 0\)
          • \(y \, dx - (x + 3y^2) \, dy = 0\)

        • 3.
          Find:

          The principal value of \[ \sec^{-1}(\sqrt{2})+2\csc^{-1}(-2) \] is:

            • \(-\frac{\pi}{2}\)
            • \(-\frac{\pi}{4}\)
            • \(\frac{\pi}{4}\)
            • \(\frac{\pi}{2}\)

          • 4.
            Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]


              • 5.

                At a birthday party, children are being served orange juice in conical cups, as shown in the figure. 


                Each cup is 15 cm deep and has a radius 5 cm. The juice is being poured into this cup at a rate of 0·1 cm3/s.
                On the basis of the above information, answer the following questions :


                  • 6.
                    Find:

                    If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

                      • \(p = -\frac{2}{3}, \, q = \frac{5}{3}\)
                      • \(p = -\frac{8}{3}, \, q = \frac{20}{3}\)
                      • \(p = \frac{20}{3}, \, q = -\frac{8}{3}\)
                      • \(p = 0, \, q = 0\)
                    CBSE CLASS XII Previous Year Papers

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