# How do you differentiate y=(sinx)^lnx?

Feb 21, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(\sin x\right)}^{\ln x} \left\{\left(\ln x\right) \left(\cot x\right) + \frac{\ln \left(\sin x\right)}{x}\right\}$

#### Explanation:

Generally when dealing with a variable exponent it is easier to differentiate (and understand) by taking natural logarithms (to remove the exponent) and differentiating implicitly:

We have:

$y = {\left(\sin x\right)}^{\ln x}$

Take Natural logarithms:

$\ln y = \ln \left\{{\left(\sin x\right)}^{\ln x}\right\}$
$\setminus \setminus \setminus \setminus \setminus \setminus = \left(\ln x\right) \left(\ln \left(\sin x\right)\right) \setminus \setminus \setminus$ (rule of logs)

Differentiate wrt $x$ (LHS implicitly; RHS product rule with chain rule):

$\setminus \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \left(\ln x\right) \left(\frac{1}{\sin} x \cdot \cos x\right) + \left(\frac{1}{x}\right) \left(\ln \left(\sin x\right)\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(\ln x\right) \left(\cot x\right) + \frac{\ln \left(\sin x\right)}{x}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = y \left\{\left(\ln x\right) \left(\cot x\right) + \frac{\ln \left(\sin x\right)}{x}\right\}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\left(\sin x\right)}^{\ln x} \left\{\left(\ln x\right) \left(\cot x\right) + \frac{\ln \left(\sin x\right)}{x}\right\}$