Ex



Misc Examples of the Chain Rule

Ex. Trigonometric functions with angle nx

(sin 5x)′ = (cos 5x)(5x)′ = (cos 5x)∙5 = 5 cos(5x)

(csc 4x)′ = (−csc 4x cot 4x)(4x)′ = −4 csc(4x) cot(4x)

In fact, for any number n,

(sin nx)′ = (cos nx)(nx)′ = (cos nx)∙n = n cos(nx)

The other trig functions have similar results, hence

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Ex. Differentiate f (x) = sin(x + sin 2x)

f ′(x) = cos(x + sin 2x)∙ (x + sin 2x)′ = cos(x + sin 2x)∙(1 + 2 cos 2x)

= (1 + 2 cos 2x) cos(x + sin 2x)

Ex. Given values of f , g, f ′, and g′ :

|x | f(x) | g(x) | f ′(x) |g′(x) |

|1 |3 |2 |4 |6 |

|2 |1 |8 |5 |7 |

|3 |7 |2 |7 |9 |

(a) If h(x) = f (g(x)), find h′(1).

By the Chain Rule, h′(x) = f ′(g(x)) g′(x). Thus, h′(1) = f ′(g(1)) g′(1) = f ′(2) g′(1) = 5∙6 = 30.

(b) If H(x) = g(f (x)), find H′(1).

Similarly, H′(x) = g′(f (x)) f ′(x). Thus, H′(1) = g′(f (1)) f ′(1)

= g′(3) f ′(1) = 9∙4 = 36.

(c) If F(x) = [h(x)]5 = [f (g(x))]5, find F ′(1).

F ′(x) = 5 [h(x)]4 h′(x) = 5 [f (g(x))]4 f ′(g(x)) g′(x).

Hence, F ′(1) = 5 [f (g(1))]4 f ′(g(1)) g′(1) = 5 [f (2)]4 f ′(2) g′(1)

= 5(1)4∙5∙6 = 150.

Ex. Let r(x) = f (g(h(x))), where h(1) = 2, g(2) = 3, h′(1) = 4, g′(2) = 5, and

f ′(3) = 6. Find r ′(1).

By the Chain Rule, r ′(x) = f ′(g(h(x))) g′(h(x)) h′(x).

Therefore, r ′(1) = f ′(g(h(1))) g′(h(1)) h′(1) = f ′(g(2)) g′(2) h′(1)

= f ′(3) g′(2) h′(1) = 6∙5∙4 = 120.

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