# How do you differentiate y=(sin(3x))^(ln(x))?

Sep 29, 2017

$y ' = {\left(\sin 3 x\right)}^{\ln x} \cdot \left(\left(\ln x\right) \left(3 \cot 3 x\right) + \frac{\ln \left(\sin 3 x\right)}{x}\right)$

#### Explanation:

One way you can do this is by using logarithms in the beginning to rewrite the formula without the exponent, and then proceed with implicit differentiation. This will require you to substitute back at the end to eliminate $y$ from your derivative result.

Begin by taking the log of both sides of the equation:

$\ln y = \ln \left({\left(\sin 3 x\right)}^{\ln x}\right)$
$\ln y = \left(\ln x\right) \left(\ln \left(\sin 3 x\right)\right) \text{ } \left[A\right]$

In [A] we used the logarithm rule $\ln {a}^{b} = b \ln a$.

Now we perform implicit differentiation. For brevity, I will use the notation $y '$ to represent $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\frac{1}{y} y ' = \left(\ln x\right) \left(\frac{1}{\sin 3 x} \cdot \left(\cos 3 x\right) \cdot 3\right) + \left(\ln \left(\sin 3 x\right)\right) \left(\frac{1}{x}\right) \text{ } \left[B\right]$

$\frac{1}{y} y ' = \left(\ln x\right) \left(3 \cot 3 x\right) + \frac{\ln \left(\sin 3 x\right)}{x}$

$y ' = y \cdot \left(\left(\ln x\right) \left(3 \cot 3 x\right) + \frac{\ln \left(\sin 3 x\right)}{x}\right)$
$y ' = {\left(\sin 3 x\right)}^{\ln x} \cdot \left(\left(\ln x\right) \left(3 \cot 3 x\right) + \frac{\ln \left(\sin 3 x\right)}{x}\right) \text{ } \left[C\right]$

Notes:

In [B] the Product Rule is combined with the Chain Rule to determine the derivative of the right side.

In [C] the original equation is substituted for $y$ in the right hand side in order to get a full derivative in terms of $x$ only.