Question

find the derivative of
$y = \left(\frac{x^{2}+7}{8}\
ight)^{8}$
$\frac{dy}{dx}=$

Explanation:

Step1: Apply chain - rule

Let $u=\frac{x^{2}+7}{8}$, then $y = u^{8}$. The chain - rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$.

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

Differentiate $y = u^{8}$ with respect to $u$. Using the power rule $\frac{d}{du}(u^{n})=nu^{n - 1}$, we get $\frac{dy}{du}=8u^{7}$.

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

Differentiate $u=\frac{x^{2}+7}{8}=\frac{1}{8}x^{2}+\frac{7}{8}$ with respect to $x$. Using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $\frac{du}{dx}=\frac{1}{8}\cdot2x=\frac{x}{4}$.

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

Substitute $\frac{dy}{du}$ and $\frac{du}{dx}$ into the chain - rule formula: $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=8u^{7}\cdot\frac{x}{4}$. Replace $u=\frac{x^{2}+7}{8}$ back in, we get $\frac{dy}{dx}=8(\frac{x^{2}+7}{8})^{7}\cdot\frac{x}{4}=\frac{x(x^{2}+7)^{7}}{4}$.

Answer:

$\frac{x(x^{2}+7)^{7}}{4}$