# How do you find the derivative for y=(sin(3x))^(ln(x))?

Use logarithmic differentiation. Take the natural logarithm of both sides of $y = {\left(\sin \left(3 x\right)\right)}^{\ln \left(x\right)}$ to get $\ln \left(y\right) = \ln \left({\left(\sin \left(3 x\right)\right)}^{\ln \left(x\right)}\right)$. Then use the property $\ln \left({a}^{b}\right) = b \ln \left(a\right)$ (for $a > 0$) to write $\ln \left(y\right) = \ln \left(x\right) \ln \left(\sin \left(3 x\right)\right)$.
Now differentiate both sides of this last equation with respect to $x$, keeping in mind that $y$ is a function of $x$ and using the Chain Rule and the Product Rule:
$\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{x} \cdot \ln \left(\sin \left(3 x\right)\right) + \ln \left(x\right) \cdot \frac{1}{\sin \left(3 x\right)} \cdot 3 \cos \left(3 x\right) .$
Simplifying a bit and multiplying both sides of this last equation by $y = {\left(\sin \left(3 x\right)\right)}^{\ln \left(x\right)}$ helps us conclude that
$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(\sin \left(3 x\right)\right)}^{\ln \left(x\right)} \cdot \left(\setminus \frac{\ln \left(\sin \left(3 x\right)\right)}{x} + 3 \ln \left(x\right) \cdot \cot \left(3 x\right)\right)$