Question

example 2 differentiate (a) ( y = sin(x^5) ) and (b) ( sin^5(x) ).
solution
(a) if ( y = sin(x^5) ), then the outer function is the sine function and the inner function is the power function, so the chain rule gives
( \frac{dy}{dx} = \frac{d}{dx} sin left( x^5
ight) = cos left( x^5
ight) cdot \frac{d}{dx} x^5 ).

(b) note that ( sin^5(x) = (sin(x))^5 ). here the outer function is the power function and the inner function is the sine function. so
( \frac{dy}{dx} = \frac{d}{dx} (sin(x))^5 = 5 cdot (sin(x))^4 cdot \frac{d}{dx} sin(x) ).

Explanation:

Step1: Identify outer and inner functions for (a)

Outer function: $\sin(u)$ (where $u=x^5$), inner function: $x^5$.
Derivative of outer at inner: $\cos(x^5)$.
Derivative of inner: $5x^4$.
$\frac{dy}{dx} = \cos(x^5) \cdot 5x^4 = 5x^4\cos(x^5)$.

Step2: Identify outer and inner functions for (b)

Outer function: $u^5$ (where $u=\sin(x)$), inner function: $\sin(x)$.
Derivative of outer at inner: $5(\sin(x))^4$.
Derivative of inner: $\cos(x)$.
$\frac{dy}{dx} = 5(\sin(x))^4 \cdot \cos(x) = 5\sin^4(x)\cos(x)$.

Answer:

For (a): $5x^4\cos(x^5)$
For (b): $5\sin^4(x)\cos(x)$