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:

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

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