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

3 Answers
Jul 27, 2015

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

Jul 27, 2015

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.

Aug 14, 2017

More generally...

Explanation:

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