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

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)#