What is the logarithmic differentiation to evaluate for f(x)=(x)^ln(3) ?

Mar 3, 2018

$f ' \left(x\right) = \ln \left(3\right) {x}^{\ln \left(3\right) - 1}$

Explanation:

We have: $f \left(x\right) = {x}^{\ln \left(3\right)} R i g h t a r r o w y = {x}^{\ln \left(3\right)}$

Let's apply $\ln$ to both sides of the equation:

$R i g h t a r r o w \ln \left(y\right) = \ln \left({x}^{\ln \left(3\right)}\right)$

$R i g h t a r r o w \ln \left(y\right) = \ln \left(3\right) \cdot \ln \left(x\right)$

Then, let's differentiate both sides with respect to $x$:

$R i g h t a r r o w \frac{d}{\mathrm{dx}} \left(\ln \left(y\right)\right) = \frac{d}{\mathrm{dx}} \left(\ln \left(3\right) \cdot \ln \left(x\right)\right)$

$R i g h t a r r o w \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{d}{\mathrm{dy}} \left(\ln \left(y\right)\right) = \ln \left(3\right) \frac{d}{\mathrm{dx}} \left(\ln \left(x\right)\right)$

$R i g h t a r r o w \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{1}{y} = \frac{\ln \left(3\right)}{x}$

$R i g h t a r r o w \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \ln \left(3\right)}{x}$

$R i g h t a r r o w \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{\ln \left(3\right)} \cdot \ln \left(3\right)}{x}$

$R i g h t a r r o w \frac{\mathrm{dy}}{\mathrm{dx}} = \ln \left(3\right) {x}^{\ln \left(3\right) - 1}$

$\therefore f ' \left(x\right) = \ln \left(3\right) {x}^{\ln \left(3\right) - 1}$