The Chain Rule is a differentiation rule used to find the derivative of a composite function, i.e., a function inside another function.
Solved Examples
Listed below are a few solved examples on Chain Rule to enhance your understanding of the concept:
Example 1: Solve, y(x) = (2x2+ 8)2
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).
Example 2: Solve, y(x) = Cos(4x).
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).
Example 3: Solve, y(x) = ln(x2 - 1).
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).
Example 4: Solve, y(x) = (ln x)2.
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).
Example 5: Solve, y(x) = √(x3 + 56).
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.
Example 6: Find the derivative of the function sin (ax+b)
Solution:
Given function is: f(x) = sin(ax+b) [it is a composite function] Differentiate with respect to x, d/dx (f(x)) = d/dx(sin(ax+b)) By the chain rule formula, dy/dx = dy/du · du/dx f''(x)= d sin(ax+b)/d(ax+b) × d(ax+b)/ dx f'(x)= cos(ax+b) × [ d(ax)/dx + d(b)/dx] = cos(ax+b) × [a × 1 + 0 ] =cos(ax+b) × a f'(x) =a cos(ax+b)
Example 7: Find the derivative of the function , f(x)= (3x+4)2
Solution:
Given function is: f(x)=(3x+4)2. Differentiate with respect to x, d/dx (f(x)) = d/dx (3x+4)2 By the chain rule formula, dy/dx = dy/du · du/dx f''(x)=d(3x+4)2 / d(3x+4) × d(3x+4)/ dx f'(x)= 2 (3x+4) × [d(3x)/dx + d(4)/dx] f'(x) = 2(3x+4) × [3 × 1 + 0] f'(x) = 2(3x+4) × 3 f'(x)= 6(3x+4)
Example 8: Find the derivative of the function f(x) = log(2x2 + 5)
Solution:
Given function is : f(x) = log(2x2+ 5) The given function is composite function so, we are using chain rule to solve the problem. By chain rule formula, dy/dx = dy/du . du/dx f '(x) = d(log(2x2 +5)) / d(2x2 +5) . d(2x2 +5) / dx = 1/(2x2+5) . 4x f '(x) = 4x / (2x2+5)
Example 9: Find the derivative of the function f(x) = √(6x + 5)
Solution:
Given function is: f(x) = √(6x + 5) The given function is composite function so, we are using chain rule to solve the problem. By chain rule formula, dy/dx = dy/du . du/dx f '(x) =d(√(6x + 5)) / d(6x + 5) . d(6x + 5) /dx f '(x)= 1/2 (√(6x + 5)) . 6 f '(x) =3 / √(6x + 5)
Example 10: Find dy/dx if y = 4x^3 + 2x^2 + 5x - 3?
Solution:
To find the derivative of y with respect to x, we need to take the derivative of each term separately. Using the power rule, we get: dy/dx = 12x^2 + 4x + 5 Therefore, the derivative of y with respect to x is 12x^2 + 4x + 5.
Practice Problems
1. Find the derivative of f(x) = sin(3x).
2. Calculate the derivative of f(x) = ln(2x2 + 1).
3. Find the derivative of f(x) = (4x2 + 1)5
4. Calculate the derivative of f(x) = √(3x - 1).
5. Determine the derivative of f(x) = cos(2x2 - 1).