## Chain Rule of Differentiation
Let f(x) = (g _{o} h)(x) = g(h(x))
Let u = h(x)
Using the above, function f may be written as:
f(x) = g(u)
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 1 Let 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 2 Let u = x^{ 3} - 4x + 5 and f(u) = u^{ 4} , hence
du / 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 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) = √ (x**^{ 2} + 2x -1)### Solution to Example 3 Let u = x^{ 2} + 2x -1 and f(u) = √u , hence
du / 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 obtain
f '(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 4 Let u = sin (2x + 3) and f(u) = u^{ 2} , hence
du / 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 5 Let u = x^{2} + x and f(u) = ln u , hence
du / 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 exercises
1 ) 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 derivatives
Solve Rate of Change Problems in Calculus
Solve Tangent Lines Problems in Calculus
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