find the derivative. f(x) if f(x)=(e^x^4 + 1)^7 f(x)=

Explanation:

Step1: Apply chain - rule

Let $u = e^{x^{4}}+1$, then $F(x)=u^{7}$. By the chain - rule $\frac{dF}{dx}=\frac{dF}{du}\cdot\frac{du}{dx}$. First, find $\frac{dF}{du}$. Using the power rule $\frac{d}{du}(u^{n})=nu^{n - 1}$, we have $\frac{dF}{du}=7u^{6}$.

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

Let $v = x^{4}$, then $u = e^{v}+1$. By the chain - rule $\frac{du}{dx}=\frac{du}{dv}\cdot\frac{dv}{dx}$. We know that $\frac{du}{dv}=e^{v}$ and $\frac{dv}{dx}=4x^{3}$. Substituting $v = x^{4}$ back, $\frac{du}{dx}=e^{x^{4}}\cdot4x^{3}$.

Step3: Calculate $\frac{dF}{dx}$

Since $\frac{dF}{dx}=\frac{dF}{du}\cdot\frac{du}{dx}$, substitute $\frac{dF}{du}=7u^{6}$ and $\frac{du}{dx}=4x^{3}e^{x^{4}}$ and $u = e^{x^{4}}+1$ back. We get $F^{\prime}(x)=7(e^{x^{4}} + 1)^{6}\cdot4x^{3}e^{x^{4}}=28x^{3}e^{x^{4}}(e^{x^{4}} + 1)^{6}$.

Answer:

$28x^{3}e^{x^{4}}(e^{x^{4}} + 1)^{6}$