Question

What is the derivative of `y=(sinx)^x`?

Answer 1

`dy/dx = (ln(sinx)+xcotx)(sinx)^x`

Explanation:

Use logarithmic differentiation.

`y = (sinx)^x`

`lny = ln((sinx)^x) = xln(sinx)` (Use properties of `ln`)

Differentiate implicitely: (Use the product rule and the chain ruel)

`1/y dy/dx = 1ln(sinx) + x [1/sinx cosx]`

So, we have:

`1/y dy/dx = ln(sinx) + x cotx`

Solve for `dy/dx` by multiplying by `y = (sinx)^x`,

`dy/dx = (ln(sinx)+xcotx)(sinx)^x`

Answer 2

`d/dx(sinx)^x=(ln(sinx)+xcotx)(sinx)^x`

Explanation:

The easiest way to see this is using:

`(sinx)^x=e^(ln((sinx)^x))=e^(xln(sinx))`

Taking the derivative of this gives:

`d/dx(sinx)^x=(d/dxxln(sinx))e^(xln(sinx))`

`=(ln(sinx)+xd/dx(ln(sinx)))(sinx)^x`

`=(ln(sinx)+x(d/dxsinx)/sinx)(sinx)^x`

`=(ln(sinx)+xcosx/sinx)(sinx)^x`

`=(ln(sinx)+xcotx)(sinx)^x`

Now we must note that if `(sinx)^x=0`, `ln((sinx)^x)` is undefined.

However, when we analyse the behaviour of the function around the `x`'s for which this holds, we find that the function behaves well enough for this to work, because, if:

`(sinx)^x` approaches 0

then:

`ln((sinx)^x)` will approach `-oo`

so:

`e^(ln((sinx)^x))` will approach 0 as well

Furthermore, we note that if `sinx<0`, `ln((sinx)^x)` will be a complex number; however, all the algebra and calculus that we have used work in the complex plane as well, so this is not a problem.

Answer 3

More generally...

Explanation:

`d/dx [f(x)^g(x)] = [g(x)/f(x)f'(x) + g'(x)ln(f(x))][f(x)^g(x)]`