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.




Web www.analyzemath.com

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



Home Page -- HTML5 Math Applets for Mobile Learning -- Math Formulas for Mobile Learning -- Algebra Questions -- Math Worksheets -- Free Compass Math tests Practice
Free Practice for SAT, ACT Math tests -- GRE practice -- GMAT practice Precalculus Tutorials -- Precalculus Questions and Problems -- Precalculus Applets -- Equations, Systems and Inequalities -- Online Calculators -- Graphing -- Trigonometry -- Trigonometry Worsheets -- Geometry Tutorials -- Geometry Calculators -- Geometry Worksheets -- Calculus Tutorials -- Calculus Questions -- Calculus Worksheets -- Applied Math -- Antennas -- Math Software -- Elementary Statistics High School Math -- Middle School Math -- Primary Math
Math Videos From Analyzemath
Author - e-mail


Updated: 2 April 2013

Copyright © 2003 - 2014 - All rights reserved