How to solve this derivative. f(x)={ln(3x)}^2x with steps...

Mathematics

1 answer:

AysviL [449]3 years ago

8 0

Rewrite <em>f(x)</em> 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,

$f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + 2x \cdot \dfrac1{\ln(3x)} \cdot \left(\ln(3x)\right)' \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + \dfrac{2x}{\ln(3x)} \cdot \dfrac1{3x} \cdot (3x)' \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + \dfrac{2}{3\ln(3x)} \cdot 3 \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + \dfrac{2}{\ln(3x)} \right)$

Then simplify this to

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

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