find the derivative. f(x) if f(x)=(e^x^8 + 1)^7 f(x)=

Explanation:

Step1: Apply chain - rule

Let $u = e^{x^{8}}+1$, then $F(x)=u^{7}$. The chain - rule states that $F^\prime(x)=\frac{dF}{du}\cdot\frac{du}{dx}$. First, find $\frac{dF}{du}$. Using the power rule $\frac{d}{du}(u^{n})=nu^{n - 1}$, we have $\frac{dF}{du}=7u^{6}=7(e^{x^{8}} + 1)^{6}$.

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

Let $v=x^{8}$, then $u = e^{v}+1$. First, find $\frac{du}{dv}$ and $\frac{dv}{dx}$. $\frac{du}{dv}=e^{v}=e^{x^{8}}$ and $\frac{dv}{dx}=8x^{7}$ by the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ and the derivative of the exponential function $\frac{d}{dv}(e^{v})=e^{v}$. Then, by the chain - rule $\frac{du}{dx}=\frac{du}{dv}\cdot\frac{dv}{dx}=e^{x^{8}}\cdot8x^{7}$.

Step3: Calculate $F^\prime(x)$

By the chain - rule $F^\prime(x)=\frac{dF}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dF}{du}=7(e^{x^{8}} + 1)^{6}$ and $\frac{du}{dx}=8x^{7}e^{x^{8}}$ into the formula, we get $F^\prime(x)=7(e^{x^{8}} + 1)^{6}\cdot8x^{7}e^{x^{8}}=56x^{7}e^{x^{8}}(e^{x^{8}} + 1)^{6}$.

Answer:

$56x^{7}e^{x^{8}}(e^{x^{8}} + 1)^{6}$