Question

find the derivative of the given function.
y = x^2 sin^9 x + x cos^(-3) x
\frac{dy}{dx}=square

Explanation:

Step1: Apply sum - rule of derivatives

The derivative of a sum of functions $u(x)+v(x)$ is $u^{\prime}(x)+v^{\prime}(x)$. Let $u(x)=x^{2}\sin^{9}x$ and $v(x)=x\cos^{- 3}x$. So, $\frac{dy}{dx}=\frac{d}{dx}(x^{2}\sin^{9}x)+\frac{d}{dx}(x\cos^{-3}x)$.

Step2: Apply product - rule on $u(x)$

The product - rule states that if $y = f(x)g(x)$, then $y^{\prime}=f^{\prime}(x)g(x)+f(x)g^{\prime}(x)$. For $u(x)=x^{2}\sin^{9}x$, let $f(x)=x^{2}$ and $g(x)=\sin^{9}x$. Then $f^{\prime}(x) = 2x$. To find $g^{\prime}(x)$, use the chain - rule. Let $u=\sin x$, so $g(u)=u^{9}$. Then $\frac{dg}{du}=9u^{8}$ and $\frac{du}{dx}=\cos x$. By the chain - rule, $g^{\prime}(x)=\frac{dg}{du}\cdot\frac{du}{dx}=9\sin^{8}x\cos x$. So, $\frac{d}{dx}(x^{2}\sin^{9}x)=2x\sin^{9}x + 9x^{2}\sin^{8}x\cos x$.

Step3: Apply product - rule on $v(x)$

For $v(x)=x\cos^{-3}x$, let $f(x)=x$ and $g(x)=\cos^{-3}x$. Then $f^{\prime}(x)=1$. To find $g^{\prime}(x)$, use the chain - rule. Let $u = \cos x$, so $g(u)=u^{-3}$. Then $\frac{dg}{du}=-3u^{-4}$ and $\frac{du}{dx}=-\sin x$. By the chain - rule, $g^{\prime}(x)=\frac{dg}{du}\cdot\frac{du}{dx}=3\cos^{-4}x\sin x$. So, $\frac{d}{dx}(x\cos^{-3}x)=\cos^{-3}x+3x\cos^{-4}x\sin x$.

Step4: Combine the results

$\frac{dy}{dx}=2x\sin^{9}x + 9x^{2}\sin^{8}x\cos x+\cos^{-3}x+3x\cos^{-4}x\sin x$.

Answer:

$2x\sin^{9}x + 9x^{2}\sin^{8}x\cos x+\cos^{-3}x+3x\cos^{-4}x\sin x$