How to differentiate x^(cosx) = y^(sinx) with respect to x

[styxrihocc]

Homework Statement

Differentiate x^(cosx) = y ^(sinx) with respect to x

Homework Equations

The Attempt at a Solution

I tried using natural logs but I am not sure if its correct, if it's wrong please point me to the right direction, thanks

x^(cosx) = y^(sinx)
ln x^(cosx) = ln y ^(sinx)
ln x (cosx) = ln y (sinx)
cosx/x - sinx lnx = cosx lny +sinx/y (dy/dx)
cosx/x - sinx lnx - cosx lny = sinx/y (dy/dx)
(cosx/x - sinx lnx - cosx lny) / (sinx/y) = dy/dx

 

[LCKurtz]

Homework Statement

Differentiate x^(cosx) = y ^(sinx) with respect to x

Homework Equations

The Attempt at a Solution

I tried using natural logs but I am not sure if its correct, if it's wrong please point me to the right direction, thanks

x^(cosx) = y^(sinx)
ln x^(cosx) = ln y ^(sinx)
ln x (cosx) = ln y (sinx)
cosx/x - sinx lnx = cosx lny +sinx/y (dy/dx)
cosx/x - sinx lnx - cosx lny = sinx/y (dy/dx)
(cosx/x - sinx lnx - cosx lny) / (sinx/y) = dy/dx

Your work is correct. You could be slightly less ambiguous with parentheses though.

 

[styxrihocc]

Thanks.