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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
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