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 the derivative of f with respect to x, f ' is given by: f '(x) = (df / 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 1Let u = 5x - 2 and f(u) = 4 cos u, hence du / 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 2: Find the derivative f '(x), if f is given by
f(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}Use the chain rule 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 3: Find f '(x), if f is given by
f(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) Use the chain rule 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 4: Find the first derivative of f if f is given by
f(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 Use the chain rule 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 5: Find the first derivative of f if f is given by
f(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: 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 on differentiation and derivatives Solve Rate of Change Problems in Calculus Solve Tangent Lines Problems in Calculus Home Page |