How do you differentiate #y=(sinx)^(x)#?

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Answer 1 · 2015-03-30T10:33:47

The answer is: #y'=e^(xlnsinx)(lnsinx+xcotx)#.

Remembering the exponential-logarithmic formula:

#[f(x)]^g(x)=e^(ln[f(x)]^g(x))=e^(g(x)lnf(x)#,

than our functio becomes:

#y=e^(xlnsinx)#,

and so, for the chain rule:

#y'=e^(xlnsinx)*(1*lnsinx+x*1/sinx*cosx)=#

#=e^(xlnsinx)(lnsinx+xcotx)#,

or, if you want:

#y'=(sinx)^x(lnsinx+xcotx)#.