# 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

## Chain Rule of DifferentiationLet f(x) = (g_{o} h)(x) = g(h(x))
Let u = h(x) Using the above, function f may be written as: the derivative of f with respect to x, f ' is given by: f '(x) = (df / du) (du / dx)## Examples Using the Chain Rule of DifferentiationWe now present several examples of applications of the chain rule.## Example 1Find the derivative f '(x), if f is given byf(x) = 4 cos (5x - 2)## Solution to Example 1Let u = 5x - 2 and f(u) = 4 cos u, hencedu / dx = 5 and df / du = - 4 sin u We now use the chain rule f '(x) = (df / 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 2Find the derivative f '(x), if f is given byf(x) = (x^{ 3} - 4x + 5)^{ 4}## Solution to Example 2Let u = x^{ 3} - 4x + 5 and f(u) = u^{ 4} , hencedu / dx = 3 x ^{ 2} - 4 and df / du = 4 u^{3}The use of the chain rule gives f '(x) = (df / du) (du / dx) = (4 u ^{ 3}) (3 x^{ 2} - 4)
We now substitute u = x ^{ 3} - 4x + 5 above to obtainf '(x) = 4 (x ^{ 3} - 4x + 5)^{ 3} (3 x^{ 2} - 4)
## Example 3Find f '(x), if f is given byf(x) = √ (x^{ 2} + 2x -1)## Solution to Example 3Let u = x^{ 2} + 2x -1 and f(u) = √u , hencedu / dx = 2x + 2 and df / du = 1 / (2 √u) Using the chain rule we obtain f '(x) = (df / du) (du / dx) = () 1 / (2 √u) ) (2x + 2) Substitute u = x ^{ 2} + 2x -1 above to obtainf '(x) = (2x + 2) ( 1 / (2 √(x ^{ 2} + 2x -1)) )
Factor 2 in numerator and denominator and simplify f '(x) = (x + 1) / (√(x ^{ 2} + 2x -1))
## Example 4Find the first derivative of f if f is given byf(x) = sin ^{ 2} (2x + 3)## Solution to Example 4Let u = sin (2x + 3) and f(u) = u^{ 2} , hencedu / dx = 2 cos(2x + 3) and df / du = 2 u The use of the chain rule leads to f '(x) = (df / 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 formula sin (2x) = 2 sin x cos x to simplify f '(x) f '(x) = 2 sin (4x + 6) ## Example 5Find the first derivative of f if f is given byf(x) = ln(x^{2} + x)## Solution to Example 5Let u = x^{2} + x and f(u) = ln u , hencedu / dx = 2 x + 1 and df / du = 1 / u Use the chain rule and substitute f '(x) = (df / du) (du / dx) = (1 / u) (2x + 1) = (2x + 1) /(x ^{2} + x)
## Exercises On Chain RuleUse the chain rule to find the first derivative to each of the functions. 1) f(x) = cos (3x -3)2) l(x) = (3x ^{ 2} - 3 x + 8)^{ 4}3) m(x) = sin [ 1 / (x - 2)] 4) t(x) = √ (3x ^{ 2} - 3 x + 6)
5) r(x) = sin ^{ 2} (4 x + 20)
## Solutions to the above exercises1 ) f '(x) = -3 sin (3 x - 3)2 ) l(x) = 12 (2 x - 1) (3 x ^{ 2} - 3 x + 8)^{ 3}3 ) m(x) = - 1 / (x - 2) ^{ 2} cos [ 1 / (x - 2)]
4 ) t(x) = (3/2)(2 x - 1) / √ (3 x ^{ 2} - 3 x + 6)
5 ) r(x) = 4 sin (8x + 4) ## More Links and Referencesdifferentiation and derivativesSolve Rate of Change Problems in Calculus Solve Tangent Lines Problems in Calculus Home Page |