differentiate y = 7^x^4 + 9 d/dx (7^x^4 + 9) =

Explanation:

Step1: Recall derivative rules

We use the sum - rule of differentiation $\frac{d}{dx}(u + v)=\frac{d}{dx}u+\frac{d}{dx}v$ where $u = 7^{x^{4}}$ and $v = 9$. Also, for a constant $c$, $\frac{d}{dx}(c)=0$, so $\frac{d}{dx}(9)=0$. For $y = a^{u}$ (where $a>0,a\neq1$ and $u$ is a function of $x$), the chain - rule gives $\frac{dy}{dx}=a^{u}\ln a\cdot\frac{du}{dx}$. Here $a = 7$ and $u=x^{4}$.

Step2: Differentiate $7^{x^{4}}$

Let $u = x^{4}$. Then $\frac{d}{dx}(7^{x^{4}})=7^{x^{4}}\ln 7\cdot\frac{d}{dx}(x^{4})$. Since $\frac{d}{dx}(x^{n})=nx^{n - 1}$ and $n = 4$, $\frac{d}{dx}(x^{4})=4x^{3}$. So $\frac{d}{dx}(7^{x^{4}})=7^{x^{4}}\ln 7\cdot4x^{3}=4x^{3}\ln 7\cdot7^{x^{4}}$.

Step3: Apply sum - rule

$\frac{d}{dx}(7^{x^{4}}+9)=\frac{d}{dx}(7^{x^{4}})+\frac{d}{dx}(9)$. Since $\frac{d}{dx}(9) = 0$ and $\frac{d}{dx}(7^{x^{4}})=4x^{3}\ln 7\cdot7^{x^{4}}$, we have $\frac{d}{dx}(7^{x^{4}}+9)=4x^{3}\ln 7\cdot7^{x^{4}}$.

Answer:

$4x^{3}\ln 7\cdot7^{x^{4}}$