find the derivative of w = √((x²·4^x)^5) dw/dx =

Explanation:

Step1: Simplify the function

First, rewrite $w = \sqrt{(x^{2}\cdot4^{x})^{5}}=(x^{2}\cdot4^{x})^{\frac{5}{2}}=x^{5}\cdot4^{\frac{5}{2}x}=x^{5}\cdot32^{x}$.

Step2: Use the product - rule

The product - rule states that if $y = u\cdot v$, where $u$ and $v$ are functions of $x$, then $y^\prime=u^\prime v + uv^\prime$. Let $u = x^{5}$ and $v = 32^{x}$. The derivative of $u=x^{5}$ is $u^\prime = 5x^{4}$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$. The derivative of $v = 32^{x}$ is $v^\prime=32^{x}\ln(32)$ using the rule $\frac{d}{dx}(a^{x})=a^{x}\ln(a)$.

Step3: Calculate the derivative of $w$

By the product - rule, $\frac{dw}{dx}=u^\prime v+uv^\prime=5x^{4}\cdot32^{x}+x^{5}\cdot32^{x}\ln(32)=x^{4}\cdot32^{x}(5 + x\ln(32))$.

Answer:

$x^{4}\cdot32^{x}(5 + x\ln(32))$