Question

What is the derivative of (ln (7x))^(cos (5x))

Answer 1

See photo for solution

Answer 2

$[ln(7x)]^{cos(5x)} = e^{ln^{[ln(7x)]^{cos(5x)}}}$
$= e^{cos(5x) ln(^{ln(7x)})}$

$\text{Let u = } cos(5x) \cdot ln(ln(7x))$

$[ln(7x)]^{cos(5x)} = e^{u}$
$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$
$= e^{u} \cdot \frac{d}{dx}[cos(5x) \cdot ln(ln(7x))]$
$= e^{u} \cdot (-5sin(5x) \cdot ln(ln(7x)) + cos(5x) \cdot \frac{d}{dx}[ln(ln(7x))])$

$\text{Let v = ln(7x)}$
$\frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{dv}{dx}$
$= \frac{1}{v} \cdot \frac{1}{x}$
$= \frac{1}{xln(7x)}$

$\therefore \frac{dy}{dx} = e^{u}[-5sin(5x) \cdot ln(ln(7x)) + cos(5x) \cdot \frac{1}{xln(7x)}]$
$= [ln(7x)]^{cos(5x)} \cdot (\frac{cos(5x)}{xln(7x)} - 5ln(ln(7x))sin(5x))$