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)]#