Chain Rule Derivative Solved Examples
Listed below are a few solved examples on Chain Rule to enhance your understanding of the concept:
Chain Rule Example 1: Solve, y(x) = (2x2+ 8)2
Solution:
Here, as you can see that y(x) is a composite function.
So it can be written as f(g(x)). The outer function f(g(x)) is g(x)² and the inner function g(x) is 2x2 + 8.
So, f'(g(x)) = 2g(x), here g(x) = 2x²+ 8.
Therefore,
f'(g(x)) = 2(2x2 + 8) and g'(x) = 4x.
Now
y'(x) = f'(g(x)).g'(x)
= 2(2x2 + 8)(4x)
= 16x(x2 + 4).
Chain Rule Example 2: Solve, y(x) = Cos(4x).
Solution:
Here, as we know from earlier, y(x) is a composite function.
So it can be written as f(g(x)). The outer function f(g(x)) is Cos(g(x)) and the inner function g(x) is 4x.
So, f'(g(x)) = -Sin(g(x)), here g(x) = 4x.
Therefore,
f'(g(x)) = -Sin(4x) and g'(x) = 4.
Now
y'(x) = f'(g(x)).g'(x)
= -(Sin(4x))(4)
= -4Sin(4x).
Chain Rule Example 3: Solve, y(x) = ln(x2 – 1).
Solution:
Here, as we know from earlier, y(x) is a composite function.
So it can be written as f(g(x)). The outer function f(g(x)) is ln(g(x)) and the inner function g(x) is x2 – 1.
So, f'(g(x)) = 1/(g(x)), here g(x) = x2 – 1.
Therefore,
f'(g(x)) = 1/(x2 – 1) and g'(x) = 2x.
Now
y'(x) = f'(g(x)).g'(x)
= (1/(x2 – 1))(2x)
= 2x/(x2 – 1).
Chain Rule Example 4: Solve, y(x) = (ln x)2.
Solution:
Here, as we know from earlier, y(x) is a composite function.
So it can be written as f(g(x)). The outer function f(g(x)) is (g(x))² and the inner function g(x) is ln x.
So, f'(g(x)) = 2g(x) , here g(x) = ln x.
Therefore,
f'(g(x)) = 2(ln x) and g'(x) = 1/x.
Now
y'(x) = f'(g(x)).g'(x)
= (2(ln x))(1/x)
= 2(ln x)/(x).
Chain Rule Example 5: Solve, y(x) = √(x3 + 56).
Solution:
Here, as we know from earlier, y(x) is a composite function. So it can be written as f(g(x)). The outer function f(g(x)) is √(g(x)) and the inner function g(x) is x3 + 56.
So, f'(g(x)) = (1/2)(x3 + 56)-1/2, here g(x) = x3 + 56.
Therefore,
f'(g(x)) = (1/2)(x3 + 56)-1/2 and g'(x) = 3x2
Now
y'(x) = f'(g(x)).g'(x)
= ( (1/2)(x3 + 56)-1/2) × ( 3x2)
= [(3/2) x2] / (x3 + 56)1/2.
Chain Rule Example 6: Solve, y(x) = ln √x.
Solution:
Here, as we know from earlier, y(x) is a composite function.
So it can be written as f(g(x)). The outer function f(g(x)) is ln(g(x)) and the inner function g(x) is √x.
So, f'(g(x)) = 1/g(x), here g(x) = √x.
Therefore,
f'(g(x)) = 1/(√x) and g'(x) = (1/(2√x)).
Now
y'(x) = f'(g(x)).g'(x)
= (1/(√x))(1/(2√x))
= 1/2x
Chain Rule: Theorem, Formula and Solved Examples
Chain Rule is a way to find the derivative of composite functions. It is one of the basic rules used in mathematics for solving differential problems. It helps us to find the derivative of composite functions such as (3x2 + 1)4, (sin 4x), e3x, (ln x)2, and others. Only the derivatives of composite functions are found using the chain rule. The famous German scientist, Gottfried Leibniz gave the chain rule in the early 17th century.
Let’s learn about Chain Rule formula, derivation and examples in detail below.
Table of Content
- What is Chain Rule?
- Chain Rule Theorem
- Chain Rule Steps to find the Derivative
- Chain Rule Formula
- Chain Rule Formula Proof
- Double Chain Rule of Differentiation
- Chain Rule for Partial Derivatives
- Application of Chain Rule
- Chain Rule Derivative Solved Examples
- Chain Rule Derivative Practice Problems
- Chain Rule Differential – FAQs
Contact Us