Question

Differentiate y w.r.t x in the following cases:
(A) y= ln (x ln x)

Answer 1

The equation is given by:

$y = \ln \left( x \ln (x) \right)$

Derivative of $y$ with respect to $x$ is:

$\frac{dy}{dx} = \frac{d}{dx} \left( \ln \left( x \ln (x) \right) \right)$

$\,\,\,\,\,\,\,\,= \frac{1}{x \ln (x)} \frac{d}{dx} \left( x \ln (x) \right)$ $\hspace{1 cm} \left[ \because F^{'}(x) = (f(g(x)))^{'} g^{'} (x) \, Where \, F(x) = f(g(x)) \right]$

$\,\,\,\,\,\,\,\,= \frac{1}{x \ln (x) } \left[ x \frac{d}{dx} (\ln (x)) + \ln (x) \frac{d}{dx} (x) \right]$ $\hspace{1 cm} \left[ \because \frac{d}{dx} (f(x)g(x)) = f(x) \frac{d}{dx} (g(x)) + g(x) \frac{d}{dx} (f(x)) \right]$

$\,\,\,\,\,\,\,\,= \frac{1}{x \ln (x)} \left[ x \cdot \frac{1}{x} + \ln (x) \cdot 1 \right]$ $\hspace{1 cm} \left[ \because \frac{d}{dx} (\ln (x)) = \frac{1}{x} \, and \, \frac{d}{dx} (x^n) = nx^{n-1} \right]$

$\,\,\,\,\,\,\,\,= \frac{\ln (x) + 1}{x \ln (x)}$