Derivative of $4x^{5x}$

Question

I'm studying for an exam and I'm confused on these type of derivative problems. I know the answer I'm just confused as how to get to the answer. Would anyone mind going through the steps:

Question: Derivative of $4x^{5x}$

Answer: $4x^{5x}(5\log x +5)$.

Answer 1

For $x \neq 0$ we have $$ 4x^{5x} = 4\exp (5x \log x); $$ for $x > 0$ we have $$ D\exp (5x \log x) = \exp (5x \log x)\cdot ( 5\log x + 5) $$ by chain rule.

Can you continue from here?

Answer 2

Consider the problem of finding the derivative of $y=x^x$. (Anything else is then just chain rule.)

$y=x^x$

$\ln y = \ln x^x$

$\ln y = x \ln x$

$y = e^{x\ln x}$

So $y'=e^{x\ln x}\cdot(x\cdot \frac{1}{x}+1\cdot\ln x)$

Hence $y'=x^x\cdot(1+\ln x)$

Alternately (via implicit differentiation):

$y=x^x$

$\ln y = \ln x^x$

$\ln y = x\ln x$

Differentiating with respect to $x$:

$\frac{y'}{y} = x\cdot \frac{1}{x}+1\cdot\ln x$

$\frac{y'}{x^x} = 1+\ln x$

$y'=x^x\cdot(1+\ln x)$

Answer 3

$$\begin{align*} \frac{d}{dx} 4x^{5x} &= 4 \frac{d}{dx}e^{5x\log x}\\ &= e^{5x\log x} \frac{d}{dx}(5x\log x) \text{ Chain rule}\\ &= 4x^{5x} (5x\frac{d}{dx}\log x + 5\log x) \text{ Leibniz rule}\\ &= 4x^{5x} (5x\frac{1}{x} + 5\log x)\\ &= 4x^{5x} (5\log x + 5) \end{align*}$$

Answer 4

For this kind of problems, logarithmic differentiation makes life easy. Considering $$y=4x^{5x}$$ $$\log(y)=\log(4)+5x\log(x)$$ Differentiate both sides and get $$\frac{y'} y=5 (\log(x)+x \times \frac 1x)=5 (1+\log(x))$$

I am sure that you can take from here.