find the derivative of y with respect to x.\ny = 9x^4 arcsin(9x^4)+\\sqrt{1 - 81x^8}\n\\frac{dy}{dx}=\\square

Explanation:

Step1: Differentiate first - term using product rule

The product rule is $(uv)^\prime = u^\prime v+uv^\prime$, where $u = 9x^{4}$ and $v=\arcsin(9x^{4})$. $u^\prime=\frac{d}{dx}(9x^{4}) = 36x^{3}$, and for $v=\arcsin(9x^{4})$, using the chain - rule $\frac{d}{dx}\arcsin(u)=\frac{u^\prime}{\sqrt{1 - u^{2}}}$ with $u = 9x^{4}$, $v^\prime=\frac{36x^{3}}{\sqrt{1-(9x^{4})^{2}}}=\frac{36x^{3}}{\sqrt{1 - 81x^{8}}}$. So, $\frac{d}{dx}(9x^{4}\arcsin(9x^{4}))=36x^{3}\arcsin(9x^{4})+9x^{4}\cdot\frac{36x^{3}}{\sqrt{1 - 81x^{8}}}=36x^{3}\arcsin(9x^{4})+\frac{324x^{7}}{\sqrt{1 - 81x^{8}}}$.

Step2: Differentiate second - term using chain rule

Let $u = 1-81x^{8}$, then $y=\sqrt{u}=u^{\frac{1}{2}}$. $\frac{du}{dx}=-648x^{7}$ and $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2\sqrt{1 - 81x^{8}}}$. By the chain - rule $\frac{d}{dx}\sqrt{1 - 81x^{8}}=\frac{-648x^{7}}{2\sqrt{1 - 81x^{8}}}=\frac{-324x^{7}}{\sqrt{1 - 81x^{8}}}$.

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

$\frac{dy}{dx}=\frac{d}{dx}(9x^{4}\arcsin(9x^{4}))+\frac{d}{dx}\sqrt{1 - 81x^{8}}$. $\frac{dy}{dx}=36x^{3}\arcsin(9x^{4})+\frac{324x^{7}}{\sqrt{1 - 81x^{8}}}-\frac{324x^{7}}{\sqrt{1 - 81x^{8}}}=36x^{3}\arcsin(9x^{4})$.

Answer:

$36x^{3}\arcsin(9x^{4})$