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

Nov 6, 2016

Use implicit differentiation along with the chain and product rules.

#### Explanation:

We can find the derivative of this function implicitly. In other words, we will find the derivative of $y$, which will then allow us to find the derivative of $\sin {\left(x\right)}^{\ln \left(x\right)}$.

First, we want to get rid of the $\ln x$ exponent. We can do that by taking the natural log of both sides and using a property of logarithms, that $\ln {x}^{a}$ is equivalent to $a \ln \left(x\right)$. Thus,

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

Now, we derive both sides. For the left side, we will have the derivative of $\ln y = \frac{1}{y}$, but we can't simply say that the derivative of $y$ is 1 (using the chain rule). Rather, we say that it is $\frac{\mathrm{dy}}{\mathrm{dx}}$.

So, the left side of the equation now looks like this:

$\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

Now we take the derivative of the right side, which we can do using the product and chain rules. We get:

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

$\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \ln \frac{\sin x}{x} + \frac{\ln x \cdot \cos x}{\sin} x$

Remember that we're trying to solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$. We can do this by multiplying both sides by $y$. Thus,

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

Finally, to get our answer back into terms of $x$, we can replace the $y$ on the right side of our derivative with $\sin {x}^{\ln} x$ from the original function.

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