determine dy/dx for y = 4x^7sin^(-1)x.

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, $u = 4x^{7}$ and $v=\sin^{-1}x$.

Step2: Differentiate $u$ with respect to $x$

Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, for $u = 4x^{7}$, we have $\frac{du}{dx}=4\times7x^{6}=28x^{6}$.

Step3: Differentiate $v$ with respect to $x$

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

Step4: Substitute into product - rule

$\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}=4x^{7}\times\frac{1}{\sqrt{1 - x^{2}}}+\sin^{-1}x\times28x^{6}$. $\frac{dy}{dx}=\frac{4x^{7}}{\sqrt{1 - x^{2}}}+28x^{6}\sin^{-1}x$.

Answer:

$\frac{4x^{7}}{\sqrt{1 - x^{2}}}+28x^{6}\sin^{-1}x$