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 
