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

Explanation:

Step1: Apply chain - rule

Let $u = e^{x^{4}}+1$, 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 $\frac{d}{du}(u^{n})=nu^{n - 1}$, for $n = 7$, we have $\frac{dF}{du}=7u^{6}$.

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

Since $u = e^{x^{4}}+1$, we find the derivative of $e^{x^{4}}$ and the derivative of the constant 1 (which is 0). Let $t=x^{4}$, then $y = e^{t}$. By the chain - rule, $\frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}$. We know that $\frac{dy}{dt}=e^{t}$ and $\frac{dt}{dx}=4x^{3}$. Substituting $t = x^{4}$ back, $\frac{d}{dx}(e^{x^{4}})=e^{x^{4}}\cdot4x^{3}$. So, $\frac{du}{dx}=4x^{3}e^{x^{4}}$.

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

Using the chain - rule $F^\prime(x)=\frac{dF}{du}\cdot\frac{du}{dx}$, substitute $\frac{dF}{du}=7u^{6}$ and $\frac{du}{dx}=4x^{3}e^{x^{4}}$. Replace $u = e^{x^{4}}+1$ back into the equation. 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:

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