find the derivative. f(x) if f(x)=(e^x^4 - 5)^9 f(x)=

Explanation:

Step1: Identify the outer - inner functions

Let $u = e^{x^{4}}-5$, then $F(x)=u^{9}$.

Step2: Differentiate the outer function

Using the power - rule $\frac{d}{du}(u^{n})=nu^{n - 1}$, for $n = 9$, we have $\frac{dF}{du}=9u^{8}$.

Step3: Differentiate the inner function

Let $v=x^{4}$, then $u = e^{v}-5$. First, $\frac{du}{dv}=e^{v}$ and $\frac{dv}{dx}=4x^{3}$. By the chain - rule $\frac{du}{dx}=\frac{du}{dv}\cdot\frac{dv}{dx}$, so $\frac{du}{dx}=e^{x^{4}}\cdot4x^{3}$.

Step4: Apply the chain - rule for $F(x)$

By the chain - rule $\frac{dF}{dx}=\frac{dF}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dF}{du}=9u^{8}$ and $\frac{du}{dx}=4x^{3}e^{x^{4}}$ into it. Since $u = e^{x^{4}}-5$, we get $\frac{dF}{dx}=9(e^{x^{4}} - 5)^{8}\cdot4x^{3}e^{x^{4}}$.

Step5: Simplify the result

$F^{\prime}(x)=36x^{3}e^{x^{4}}(e^{x^{4}} - 5)^{8}$.

Answer:

$36x^{3}e^{x^{4}}(e^{x^{4}} - 5)^{8}$