Question

calculate the derivative of the following function.
y = \sqrt7{7x^{4}+\sin(9x)}
\frac{dy}{dx}=\square.

Explanation:

Step1: Rewrite the function

Rewrite $y = \sqrt[7]{7x^{4}+\sin(9x)}$ as $y=(7x^{4}+\sin(9x))^{\frac{1}{7}}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = u^{\frac{1}{7}}$ and $u = 7x^{4}+\sin(9x)$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$: $\frac{dy}{du}=\frac{1}{7}u^{-\frac{6}{7}}$.

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

Differentiate $u = 7x^{4}+\sin(9x)$ with respect to $x$. The derivative of $7x^{4}$ is $28x^{3}$ (using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$), and the derivative of $\sin(9x)$ is $9\cos(9x)$ (using the chain - rule for $\sin(u)$ where $u = 9x$ and $\frac{d}{du}\sin(u)=\cos(u)$ and $\frac{du}{dx}=9$). So, $\frac{du}{dx}=28x^{3}+9\cos(9x)$.

Step4: Calculate $\frac{dy}{dx}$

Substitute $\frac{dy}{du}$ and $\frac{du}{dx}$ into the chain - rule formula: $\frac{dy}{dx}=\frac{1}{7}(7x^{4}+\sin(9x))^{-\frac{6}{7}}\cdot(28x^{3}+9\cos(9x))=\frac{28x^{3}+9\cos(9x)}{7(7x^{4}+\sin(9x))^{\frac{6}{7}}}$.

Answer:

$\frac{28x^{3}+9\cos(9x)}{7(7x^{4}+\sin(9x))^{\frac{6}{7}}}$