What is the derivative of y=π^sinx

y=π^sinx

y = π^sinx (Take the natural logarithm of both sides)
ln(y) = ln(π^sinx)
ln(y) = sinx * ln(π) (Differentiate both sides)
1/y * dy = cosx * ln(π) dx (Isolate dy/dx)
dy = ycosx * ln(π) dx
dy/dx = ln(π) * (π^sinx) * cosx (Answer)

Have a good day.

the best answer would be through differentiation using the chain rule:

y = TT ^ sin x

y = TT ^ u

u = sinx

du = cosx

dy = u x TT

With the chain rule you have to differentiate both u and y seperately which is what I have shown above, then the final part of the chain rule is that you replace u in the dy equation with the differentiation of u.


= cosx * TT
And as TT is a constant this goes infront of the function

TTcosx

It is just analogous to a^x...Now,to ur question,if u differentiate this u'll have to apply chain rule
dy/dx=(pi)^sinx.log(pi) X cosx

where cosx is the derivative of sinx...

Hope u r satisfied with the explanation...
All the best!

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