How do I use the product rule to find dy/dx of y=(cosx)^x?

[pugola12]

Homework Statement

Find dy/dx of y=(cosx)^x

Homework Equations

d/dx of a^u=lna*a^u*u'

The Attempt at a Solution

I thougt I just had to follow the form shown above, and this is what I got.

y=(cosx)^x
dy/dx=ln(cosx)*(cosx)^x*1
dy/dx=ln(cosx)*(cosx)^x

However, the actual answer is (cosx)^x*(ln(cosx)-xtanx)
I don't understand where this comes from at all. Thanks for your input.

 

[Dick]

Homework Statement

Find dy/dx of y=(cosx)^x

Homework Equations

d/dx of a^u=lna*a^u*u'

The Attempt at a Solution

I thougt I just had to follow the form shown above, and this is what I got.

y=(cosx)^x
dy/dx=ln(cosx)*(cosx)^x*1
dy/dx=ln(cosx)*(cosx)^x

However, the actual answer is (cosx)^x*(ln(cosx)-xtanx)
I don't understand where this comes from at all. Thanks for your input.

That's not a good form to follow. It assumes that in a^u that a is a constant. That isn't true in your case. Try writing v^u=e^(log(v)*u) and differentiate that.

 

[iRaid]

ln y = x ln cosx

You can now use the product rule on the right side..
(Hint: it becomes (1/y)y' on the left)