Question

find the derivative of y with respect to x.
y = 4x^{3}\sin^{-1}(4x^{3})+sqrt{1 - 16x^{6}}
\frac{dy}{dx}=\square

Explanation:

Step1: Apply product - rule and chain - rule to $4x^{3}\sin^{- 1}(4x^{3})$

The product rule states that if $u = 4x^{3}$ and $v=\sin^{-1}(4x^{3})$, then $(uv)'=u'v + uv'$.
$u'=\frac{d}{dx}(4x^{3})=12x^{2}$, and for $v = \sin^{-1}(4x^{3})$, by the chain - rule $\frac{d}{dx}\sin^{-1}(u)=\frac{1}{\sqrt{1 - u^{2}}}\cdot u'$, where $u = 4x^{3}$ and $u' = 12x^{2}$. So $v'=\frac{12x^{2}}{\sqrt{1-(4x^{3})^{2}}}=\frac{12x^{2}}{\sqrt{1 - 16x^{6}}}$. Then $(4x^{3}\sin^{-1}(4x^{3}))'=12x^{2}\sin^{-1}(4x^{3})+4x^{3}\cdot\frac{12x^{2}}{\sqrt{1 - 16x^{6}}}$.

Step2: Apply chain - rule to $\sqrt{1 - 16x^{6}}$

Let $u = 1-16x^{6}$, then $y=\sqrt{u}=u^{\frac{1}{2}}$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dx}=-96x^{5}$. So $\frac{d}{dx}\sqrt{1 - 16x^{6}}=\frac{-96x^{5}}{2\sqrt{1 - 16x^{6}}}=\frac{-48x^{5}}{\sqrt{1 - 16x^{6}}}$.

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

$\frac{dy}{dx}=(4x^{3}\sin^{-1}(4x^{3}))'+(\sqrt{1 - 16x^{6}})'$.
$\frac{dy}{dx}=12x^{2}\sin^{-1}(4x^{3})+\frac{48x^{5}}{\sqrt{1 - 16x^{6}}}-\frac{48x^{5}}{\sqrt{1 - 16x^{6}}}$.

Answer:

$12x^{2}\sin^{-1}(4x^{3})$