Question

calculate the following derivatives using the product rule. a. $\frac{d}{dx}(sin^{2}x)$ b. $\frac{d}{dx}(sin^{3}x)$ c. $\frac{d}{dx}(sin^{4}x)$ d. based upon your answers to parts (a)-(c), make a conjecture about the value of $\frac{d}{dx}(sin^{n}x)$, where n is a positive integer.

Explanation:

Step1: Recall the product - rule for (a)

The product - rule states that if $y = u\cdot v$, then $\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$. For $y = \sin x\cdot\sin x$, let $u = \sin x$ and $v=\sin x$. Then $\frac{du}{dx}=\cos x$ and $\frac{dv}{dx}=\cos x$. So $\frac{d}{dx}(\sin^{2}x)=\sin x\cos x+\sin x\cos x = 2\sin x\cos x$.

Step2: Apply product - rule for (b)

For $y=\sin^{3}x=\sin x\cdot\sin^{2}x$. Let $u = \sin x$ and $v=\sin^{2}x$. We know $\frac{du}{dx}=\cos x$ and $\frac{dv}{dx}=2\sin x\cos x$ from part (a). Then $\frac{d}{dx}(\sin^{3}x)=\sin x\cdot(2\sin x\cos x)+\sin^{2}x\cdot\cos x=3\sin^{2}x\cos x$.

Step3: Apply product - rule for (c)

For $y = \sin^{4}x=\sin x\cdot\sin^{3}x$. Let $u=\sin x$ and $v = \sin^{3}x$. We know $\frac{du}{dx}=\cos x$ and $\frac{dv}{dx}=3\sin^{2}x\cos x$ from part (b). Then $\frac{d}{dx}(\sin^{4}x)=\sin x\cdot(3\sin^{2}x\cos x)+\sin^{3}x\cdot\cos x=4\sin^{3}x\cos x$.

Step4: Make a conjecture for (d)

Based on the above results, we can conjecture that $\frac{d}{dx}(\sin^{n}x)=n\sin^{n - 1}x\cos x$, where $n$ is a positive integer.

Answer:

a. $2\sin x\cos x$
b. $3\sin^{2}x\cos x$
c. $4\sin^{3}x\cos x$
d. $\frac{d}{dx}(\sin^{n}x)=n\sin^{n - 1}x\cos x$ (conjecture)