Question

How to solve this derivative.

f(x)={ln(3x)}^2x
with steps...

Answer 1

Rewrite f(x) as a nested exponential-logarithm expression :

`\left(\ln(3x)\right)^(2x) = \exp\left(\ln\left(\ln(3x)\right)^(2x)\right)`

(where
`\exp(x) = e^x`)

One of the properties of logarithms lets us drop the exponent as a coefficient:

`\exp\left(\ln\left(\ln(3x)\right)^(2x)\right) = \exp\left(2x\ln\left(\ln(3x)\right)\right)`

Now, by the chain rule, we have

`f(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \implies \\\\ f'(x) = \left(\exp\left(2x\ln\left(\ln(3x)\right)\right)\right)' \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2x\ln\left(\ln(3x)\right)\right)'`

By the product rule,

`f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(   (2x)' \ln\left(\ln(3x)\right) + 2x\left(\ln\left(\ln(3x)\right)\right)'   \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(   2\ln\left(\ln(3x)\right) + 2x\left(\ln\left(\ln(3x)\right)\right)'   \right)`

By the chain rule again,

`f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + 2x \cdot \frac1{\ln(3x)} \cdot \left(\ln(3x)\right)'  \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + (2x)/(\ln(3x)) \cdot \frac1{3x} \cdot (3x)' \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + (2)/(3\ln(3x)) \cdot 3 \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + (2)/(\ln(3x)) \right)`

Then simplify this to

`f'(x) = \boxed{\left(\ln(3x)\right)^(2x) \left(2\ln\left(\ln(3x)\right) + (2)/(\ln(3x)) \right)}`