# How do I find the derivative of y = (sin(e^x))^ln(x^2)?

$\textcolor{b l u e}{y ' = {\left(\sin {e}^{x}\right)}^{\ln {x}^{2}} \left[{e}^{x} \cdot \ln {x}^{2} \cdot \cot {e}^{x} + \frac{2}{x} \cdot \ln \sin {e}^{x}\right]}$

#### Explanation:

The given equation is

$y = {\left(\sin {e}^{x}\right)}^{\ln {x}^{2}}$

Take the natural logarithm of both sides then differentiate with respect to x

$y = {\left(\sin {e}^{x}\right)}^{\ln {x}^{2}}$

$\ln y = \ln {\left(\sin {e}^{x}\right)}^{\ln {x}^{2}}$

$\ln y = \left(\ln {x}^{2}\right) \cdot \ln \left(\sin {e}^{x}\right)$

We now have a derivative of a product at the right sides of the equation.

Obtain the derivative of both sides

$\ln y = \left(\ln {x}^{2}\right) \cdot \ln \left(\sin {e}^{x}\right)$

$\frac{d}{\mathrm{dx}} \left(\ln y\right) = \frac{d}{\mathrm{dx}} \left[\left(\ln {x}^{2}\right) \cdot \ln \left(\sin {e}^{x}\right)\right]$

$\frac{1}{y} \cdot y ' = \left(\ln {x}^{2}\right) \cdot \frac{d}{\mathrm{dx}} \left(\ln \sin {e}^{x}\right) + \left(\ln \sin {e}^{x}\right) \cdot \frac{d}{\mathrm{dx}} \left(\ln {x}^{2}\right)$

$\frac{1}{y} \cdot y ' = \left(\ln {x}^{2}\right) \cdot \frac{1}{\sin {e}^{x}} \cdot \frac{d}{\mathrm{dx}} \left(\sin {e}^{x}\right) + \left(\ln \sin {e}^{x}\right) \cdot \frac{1}{x} ^ 2 \cdot \frac{d}{\mathrm{dx}} \left({x}^{2}\right)$

$\frac{1}{y} \cdot y ' = \left(\ln {x}^{2}\right) \cdot \frac{1}{\sin {e}^{x}} \left(\cos {e}^{x}\right) \frac{d}{\mathrm{dx}} \left({e}^{x}\right) + \left(\ln \sin {e}^{x}\right) \cdot \frac{1}{x} ^ 2 \cdot 2 x$

$\frac{1}{y} \cdot y ' = \left(\ln {x}^{2}\right) \cdot \frac{1}{\sin {e}^{x}} \left(\cos {e}^{x}\right) \left({e}^{x}\right) + \frac{2}{x} \cdot \ln \sin {e}^{x}$

$\frac{1}{y} \cdot y ' = \left(\ln {x}^{2}\right) \cdot \left(\cot {e}^{x}\right) \left({e}^{x}\right) + \frac{2}{x} \cdot \ln \sin {e}^{x}$

Multiply both sides by $y$ then simplify

$y \cdot \frac{1}{y} \cdot y ' = y \cdot \left(\ln {x}^{2}\right) \cdot \left(\cot {e}^{x}\right) \left({e}^{x}\right) + \frac{2}{x} \cdot \ln \sin {e}^{x}$

$y ' = y \cdot \left(\ln {x}^{2}\right) \cdot \left(\cot {e}^{x}\right) \left({e}^{x}\right) + \frac{2}{x} \cdot \ln \sin {e}^{x}$

$y ' = {\left(\sin {e}^{x}\right)}^{\ln {x}^{2}} \left[{e}^{x} \cdot \ln {x}^{2} \cdot \cot {e}^{x} + \frac{2}{x} \cdot \ln \sin {e}^{x}\right]$

God bless....I hope the explanation is useful.