# How do you find the derivative of  y = (x ln(x)) / sin(x)?

Feb 13, 2016

$\frac{\sin x \left(1 + \ln x\right) - \cos x . x \ln x}{\sin} ^ 2 x$

#### Explanation:

using the$\textcolor{b l u e}{\text{ quotient rule }}$

If f(x) $= g \frac{x}{h \left(x\right)}$

then f'(x)$= \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2$

now: xlnx is a product of 2 functions and as such will require the$\textcolor{b l u e}{\text{ product rule }}$

If f(x) = g(x)h(x) then f'(x) = g(x)h'(x) + h(x)g'(x)
$\textcolor{b l a c k}{\text{-----------------------------------------------}}$
differentiating the numerator using 'product rule"

$x \frac{d}{\mathrm{dx}} \left(\ln x\right) + \ln x \frac{d}{\mathrm{dx}} \left(x\right) = x . \left(\frac{1}{x}\right) + \ln x .1 = 1 + \ln x$

differentiate the denominator: $\frac{d}{\mathrm{dx}} \left(\sin x\right) = \cos x$
$\textcolor{b l a c k}{\text{---------------------------------------------}}$

back to the original function using the 'quotient rule'

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sin x \frac{d}{\mathrm{dx}} \left(x \ln x\right) - x \ln x \frac{d}{\mathrm{dx}} \left(\sin x\right)}{\sin} ^ 2 x$

'inserting' the derivates gives

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sin x \left(1 + \ln x\right) - \cos x . x \ln x}{\sin} ^ 2 x$