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 ) ( (x^5) ) ( = ) ( cos ) ( (x^5) ) ( cdot ) ( square )
( quad ) ( overline{\text{outer}} ) ( overline{\text{evaluated}} ) ( quad ) ( overline{\text{derivative}} ) ( overline{\text{evaluated}} ) ( quad ) ( overline{\text{derivative}} )
( quad ) ( \text{function} ) ( \text{at inner} ) ( quad ) ( \text{of outer} ) ( \text{at inner} ) ( quad ) ( \text{of inner} )
( quad ) ( \text{function} ) ( \text{function} ) ( quad ) ( \text{function} ) ( \text{function} ) ( quad ) ( \text{function} )
( = ) ( square ) ( cdot )

(b) note that ( sin^5(x) = (sin(x))^square ). 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 square )
( quad ) ( overline{\text{inner}} ) ( quad ) ( overline{\text{derivative of outer}} ) ( quad ) ( overline{\text{derivative}} )
( quad ) ( \text{function} ) ( quad ) ( \text{function evaluated} ) ( quad ) ( \text{of inner} )
( quad ) ( quad ) ( quad ) ( \text{at inner function} ) ( quad ) ( \text{function} )
( = ) ( square ) ( cdot )

Explanation:

Step1: Find derivative of inner function for (a)

Inner function is $x^5$, derivative is $5x^4$.

Step2: Compute derivative for (a)

Multiply by derivative of outer function: $\cos(x^5) \cdot 5x^4$.

Step3: Identify exponent for (b)

$\sin^5(x) = (\sin(x))^5$, exponent is 5.

Step4: Find derivative of inner function for (b)

Inner function is $\sin(x)$, derivative is $\cos(x)$.

Step5: Compute derivative for (b)

Multiply by derivative of outer function: $5(\sin(x))^4 \cdot \cos(x)$.

Answer:

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