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Chain Rule Formula in differential calculus finds the derivative of the composition of two or more functions. The chain rule formula is applicable to a number of functions that make up the composition. Thus, this formula helps to calculate the derivative of a composition of functions. For example, if f and g are functions, then the chain rule expresses the derivative of their composition which is represented as d/dx [f(g(x))] = f'(g(x)) g'(x).
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Key Terms: Chain Rule Formula, Composite Functions, Differentiation, Differential calculus, Functions, Chain rule
Chain Rule Formula
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The chain rule formula for the function y = f(x), where f(x) is a composite function such that x = g(t), is given as
\(\frac{dy}{dx} = \frac{dy}{du} . \frac{du}{dx}\)
This is the standard form of differentiation formula of the chain rule.
The chain rule formula can also be represented as:
d/dx(f(g(x)) = f’(g(x))·g’(x)
\(\frac{dy}{dx} = \frac{dy}{du} . \frac{du}{dx}\)
Chain Rule Formula
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Continuity and Differentiability Detailed Video Explanation:
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Solved Example
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Here is a solved example to understand the chain rule formula better:
Example: Find the differentiation of the function, y = cosx2.
Solution: Let u=x2, then we have y = cos u
Hence: du/dx=2x and dy/du=−sin u
The chain rule says: dy/dx=dy/du.du/dx dy/dx=−sin u × 2x dy/dx= – 2xsinx2
Therefore, the derivative of y with respect to x will be (− 2xsinx2).
Hence, we can see that this method of chain rule will sometime make the difficult process of differentiation a simple computation.
Read More: Inverse Process of Differentiation
Chain Rule in Differentiation
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Let f represent a real-valued function, a composition of two functions u and v in such a manner that: f = v(u(x)) Let us take u(x) = t
Read More: Differential Equation
Now, if the functions u and v are differentiable and dt/dx and dv/dt exist, then we can say that the composite function f(x) is also differentiable. This can be performed as follows:
Using Leibnitz notation, the differentiation of the above function can be expressed as
df/dx = (dv/dt) × (dt/dx)
In order to differentiate a composite function at any point in its domain, we need to differentiate the outer part (i.e., the function enclosing some other function) first and then multiply it with the inner function’s derivative function. This will offer the desired differentiation to us.
Read More: Analytic Function
Things to Remember
- Differentiation can be defined as the process through which we can calculate the rate of change of a dependent variable in relation to a change of the independent variable.
- The chain rule helps us to determine the derivative of composite functions. The chain rule formula helps us to calculate the derivative of a composition of functions.
- The chain rule formula for the function y = f(x), where f(x) is a composite function such that x = g(t), is represented as dy/dx = dy/du.du/dx
- The chain rule formula can also be represented as d/dx(f(g(x)) = f’(g(x))·g’(x).
- The chain rule formula has made easier the calculations involving computing the derivatives of complicated expressions, such as those encountered in many physics applications.
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Sample Questions
Ques 1. Differentiate f(x) = (1 + x2)5 . (3 Marks)
Ans. Using the Chain rule, dy/dx = dy/du ⋅ du/dx 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 2. Find the derivative of the function f(x) = sin(2x2 – 6x). (3 Marks)
Ans. The given can be expressed as a composite function as given below:
f(x) = sin(2x2 – 6x) u(x) =2x2 – 6x v(t) = sin t Thus, t = u(x) = 2x2 – 6x ⇒f(x) = v(u(x))
According to the chain rule, df(x)/dx = (dv/dt) × (dt/dx) Where, dv/dt = d/dt (sin t) = cos t dt/dx = d/dx [u(x)] = d/dx (2x2 – 6x) = 4x – 6
Therefore, df/dx = cos t × (4x – 6) = cos(2x2 – 6x) × (4x – 6) = (4x – 6) cos(2x2 – 6x)
Ques 3. Find the derivative of the function given by f(x) = sin (ex3) (3 Marks)
Ans. Given, f(x) = sin (ex3) We can see that it is a composition of three functions such as:
p(s) = sin s, q(t) = et and r(x) = x3 Thus, f(x) = p(q(r(x)))
That means, t = x3 and s = ex3 Using chain rule formula, df/dx = (dp/ds) × (ds/dt) × (dt/dx) = [d/ds (sin s)] × [d/dt (et)] × [d/dx (x3)] = cos s × et × 3x2 = cosex3 x ex3 x 3x2
Ques 4. Find the derivative of the function given by f(x) = √tan(x2 + 1) (3 Marks)
Ans. Given, f(x) = √tan(x2 + 1)
The function that has been given represents a composition of functions where f(x) = √tan(x2 + 1) u(x) = x2 + 1 u(t) = √tant → f(x) = v(u(x))
According to the chain rule, df/dx = dv/dt x dt/dx → dv/dt = ½(tant)-½ x sec2 t = ½ x 1/√tant x sec2t
Also, t = u(x) → dt/dx = 2x → df/dx = 2x X ½ X 1/√tant X sec2t
Therefore, df/dx = xsec2(x2 + 1)/ √tan(x2 + 1)
Ques 5. Find the derivative of y= ln √x using the chain rule. (3 Marks)
Ans. f(x) = y is a composition of the functions.
ln(x) and √x, and therefore we can differentiate it using the chain rule.
Assume u = √x. Then y = ln u.
By the chain rule formula, dy/dx = dy/du · du/dx dy/dx = d/du (ln u) · d/dx (√x) dy/dx = (1/u) · (1/(2√x)) dy/dx = (1/√x) · (1/(2√x)) dy/dx = 1/(2x) [because u = 1/(2√x)] y = cos (2x2 + 1)
Hence, dy/dx = 1/(2x)
Ques 6. Given that a point A is moving along the curve whose equation is y = √(x3 + 56). When A is at (2,8), y is increasing at the rate of 2 units per second, now calculate how fast is x changing? (3 Marks)
Ans. To find:
dx/dt Given y = √(x3 + 56) and dy/dt = 2 units / sec dy/dx = (1/2)(x3 + 56)-1/2 (3x2) =[(3/2) x2 ] / (x3 + 56)1/2
Applying the chain rule, dx/dt = dx/dy . dy/dt
Given dy/dx at x = 2 dy/dx at x = 2is [3(4)]/2√64 dy/dx =3/4 dx/dy = 4/3
Thus, dx/dt = 4/3 X 2 = 8/3
Ques 7. What will be the derivative of the function y = cos (2x2 + 1) using the chain rule? (3 Marks)
Ans. Let us assume that, u = 2x2 + 1 Then, y = cos u
Using the chain rule formula,
dy/dx = dy/du · du/dx
dy/dx = d/du (cos u) · d/dx (2x2 + 1)
dy/dx = - sin u · 4x
dy/dx = – 4x sin (2x2 + 1) (As u = 2x2 + 1)
Therefore, the derivative of the given function is, dy/dx = -4x sin (2x2 + 1).
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