Use the Chain Rule of Differentiation in Calculus

The chain rule of differentiation of functions in calculus is presented along with several examples and detailed solutions and comments. Also in this site, Step by Step Calculator to Find Derivatives Using Chain Rule

Let f(x) = (g o h)(x) = g(h(x)) and u = g(x)

Let y = f(u); the derivative of f with respect to x, f '(x) is given by:

f '(x) = (dy / du) (du / dx)


We now present several examples of applications of the chain rule.

Example 1: Find the derivative f '(x), if f is given by

f(x) = 4 cos (5x - 2)


Solution to Example 1

  • Let u = 5x - 2 and y = 4 cos u, hence

      du / dx = 5 and dy / du = -4 sin u

  • We now use the chain rule

      f '(x) = (dy / du) (du / dx) = - 4 sin (u) (5)

  • We now substitute u = 5x - 2 in sin (u) above to obtain

      f '(x) = - 20 sin (5x - 2)


Example 2: Find the derivative f '(x), if f is given by

f(x) = (x 3 - 4x + 5) 4


Solution to Example 2

  • Let u = x 3 - 4x + 5 and y = u 4 , hence

      du / dx = 3 x 2 - 4 and dy / du = 4 u 3

  • Use the chain rule

      f '(x) = (dy / du) (du / dx) = (4 u 3) (3 x 2 - 4)

  • We now substitute u = x 3 - 4x + 5 above to obtain

      f '(x) = 4 (x 3 - 4x + 5) 3 (3 x 2 - 4)


Example 3: Find f '(x), if f is given by

f(x) = sqrt (x 2 + 2x -1)

where
sqrt means square root

Solution to Example 3

  • Let u = x 2 + 2x -1 and y = sqrt (u) , hence

      du / dx = 2x + 2 and dy / du = 1 / (2 sqrt(u))

  • Use the chain rule

      f '(x) = (dy / du) (du / dx) = [ 1 / (2 sqrt(u)) ] (2x + 2)

  • Substitute u = x 2 + 2x -1 above to obtain

      f '(x) = (2x + 2) [ 1 / (2 sqrt(x 2 + 2x -1)) ]

  • Factor 2 in numerator and denominator and simplify

      f '(x) = (x + 1) / (sqrt(x 2 + 2x -1))


Example 4: Find the first derivative of f if f is given by

f(x) = sin 2 (2x + 3)


Solution to Example 4

  • Let u = sin (2x + 3) and y = u 2 , hence

      du / dx = 2 cos(2x + 3) and dy / du = 2 u

  • Use the chain rule

      f '(x) = (dy / du) (du / dx) = 2 u 2 cos(2x + 3)

  • Substitute u = sin (2x + 3) above to obtain

      f '(x) = 4 sin (2x + 3) cos (2x + 3)

  • Use the trigonometric fromula sin (2x) = 2 sinx cos x to simplify f '(x)

      f '(x) = 2 sin (4x + 6)


Exercises: Find the first derivative to each of the functions.

1 - f(x) = cos (3x -3)

2 - l(x) = (3x
2 -3x + 8) 4

3 - m(x) = sin [ 1 / (x - 2)]

4 - t(x) = sqrt (3x
2 - 3x + 6)

5 - r(x) = sin
2 (4x + 20)

Solutions to the above exercises

1 - f '(x) = -3 sin (3x -3)

2 - l(x) = 12 (2x - 1) (3x
2 -3x + 8) 3

3 - m(x) = -1 / (x - 2)
2 cos [ 1 / (x - 2)]

4 - t(x) = (3/2)(2x - 1) / sqrt (3x
2 - 3x + 6)

5 - r(x) = 4 sin (8x + 4)


More on
differentiation and derivatives

Solve Rate of Change Problems in Calculus

Solve Tangent Lines Problems in Calculus