Question

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

Answer 1

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(x)^(ln(x))`.

First, we want to get rid of the `lnx` exponent. We can do that by taking the natural log of both sides and using a property of logarithms, that `lnx^a` is equivalent to `aln(x)`. Thus,

`lny = ln(sinx)^lnx`
`lny = lnx*ln(sinx)`

Now, we derive both sides. For the left side, we will have the derivative of `lny = 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 `dy/dx`.

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

`1/y*dy/dx`

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

`1/y*dy/dx = (ln(sinx)*1/x) + (lnx*1/sinx*cosx)`

`1/y*dy/dx = ln(sinx)/x + (lnx*cosx)/sinx`

Remember that we're trying to solve for `dy/dx`. We can do this by multiplying both sides by `y`. Thus,

`dy/dx = y(ln(sinx)/x + (lnx*cosx)/sinx)`

Finally, to get our answer back into terms of `x`, we can replace the `y` on the right side of our derivative with `sinx^lnx` from the original function.

Our final answer is:

`dy/dx = sin(x)^lnx*(ln(sinx)/x + (lnx*cosx)/sinx)`