The chain rule

Three quick questions from earlier lessons.

The violet curve is $y = \sin x$ ; the blue one is $y = \sin(kx)$ . Increase $k$ .

The curves y = sin x and y = sin(kx). With k = 2 the blue curve oscillates twice as fast, and its steepest slope is 2 instead of 1.

At $k = 2$ the blue curve goes twice as fast. At $x = 0$ , what is the slope of $y = \sin(2x)$ ?

For a function inside a function: differentiate the outside, keep the inside, multiply by the derivative of the inside.

Chain rule:

\big(f(g(x))\big)' = f'(g(x)) \cdot g'(x)

$h(x) = (x^2 + 1)^5$ . Find $h'(x)$ .

f(u) = u^5

g(x) = x^2 + 1

Outside u⁵, inside u = x² + 1.

h'(x) = 5(x^2 + 1)^4 \cdot 2x

f'(u) = 5u⁴ with the inside kept; g'(x) = 2x.

h'(x) = 10x(x^2 + 1)^4

Tidy.

Your turn: $h(x) = e^{3x}$ .

h'(x) = 3e^{3x}

Outside eᵘ, inside 3x: eᵘ times 3.

Boss problem. $h(x) = x^{ 3 }\sin(4x)$ . Find the slope of its graph at $x = \dfrac{\pi}{ 4 }$ .

Three questions to check you have got it.

Chain rule

$\big(f(g(x))\big)' = f'(g(x))\,g'(x)$ : outside's derivative (inside kept) times inside's derivative.

Watch out: do not forget the factor $g'(x)$ .

End of the lesson