question 3 of 14, step 1 of 1 determine dy/dx for y = 3x^5sin^(-1)x. answer

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$. Here, $u = 3x^{5}$ and $v=\sin^{- 1}x$.

Step2: Find $\frac{du}{dx}$

Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, for $u = 3x^{5}$, we have $\frac{du}{dx}=3\times5x^{4}=15x^{4}$.

Step3: Find $\frac{dv}{dx}$

The derivative of $\sin^{-1}x$ is $\frac{1}{\sqrt{1 - x^{2}}}$, so $\frac{dv}{dx}=\frac{1}{\sqrt{1 - x^{2}}}$.

Step4: Substitute into product - rule

$\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}=3x^{5}\times\frac{1}{\sqrt{1 - x^{2}}}+\sin^{-1}x\times15x^{4}$. $\frac{dy}{dx}=\frac{3x^{5}}{\sqrt{1 - x^{2}}}+15x^{4}\sin^{-1}x$.

Answer:

$\frac{3x^{5}}{\sqrt{1 - x^{2}}}+15x^{4}\sin^{-1}x$