Differentiate.

y= (cos x)^x

u= cos x
du= -sin x dx

ln y = ln(cos x)^x
ln y = x ln(cos x)

(dy/dx)/(y)= ln(cos x)
(dy/dx)= y ln(cos x)
= (cos x)^x * (ln cos x)

(dx/du)= x(cos x)^(x-1) * (-sin x)
= - x sin(x)cos^(x-1)(x)

(dy/dx)-(dx/du)= [(cos^x(x))(ln(cos(x)))-(x sin(x)cos^(x-1)(x)]

(dy/du)= cos^x(x)*(ln(cos(x)))-(x tan(x))

Is this correct?

Also, I am stuck on a different problem.

Differentiate.

y= arctan(arcsin(sqrt(x)))

u= arcsin(sqrt(x))
du= (1/(sqrt(1-x^2))) dx

ln y = ln ?? do I put the whole original here?

ln y = x ln(cos x) I agree.

y'/y=x/cosx + ln(cosx) which changes the rest.

check that.

I would do the next this way.

y= arctan u
y'=d(arctan u) du
now u= arcsin(z)
du= d arcsinZ dz
and z= sqrtX
dz= 1/2sqrtX dx

so do that substitution, and you are done.

I'm sorry, but I'm confused on the 2nd part?

Yes, your differentiation steps for the first problem are correct.

For the second problem, to differentiate y = arctan(arcsin(sqrt(x))), you can use the chain rule.

Let's define u = arcsin(sqrt(x)). Then, we have y = arctan(u).

Now, differentiate u = arcsin(sqrt(x)) with respect to x:
du = (1 / sqrt(1 - x^2)) * (1 / (2 * sqrt(x))) dx
du = dx / (2 * sqrt(x) * sqrt(1 - x^2))

Next, using the chain rule, differentiate y = arctan(u):
(dy / dx) = (dy / du) * (du / dx)

To find (dy / du), differentiate y = arctan(u) with respect to u:
(dy / du) = 1 / (1 + u^2)

Substituting in the values of (du / dx) and (dy / du), we have:
(dy / dx) = (1 / (1 + u^2)) * (dx / (2 * sqrt(x) * sqrt(1 - x^2)))

Since u = arcsin(sqrt(x)), substitute this back in:
(dy / dx) = (1 / (1 + (arcsin(sqrt(x)))^2)) * (dx / (2 * sqrt(x) * sqrt(1 - x^2)))

Simplifying further, you can substitute arcsin^2(u) = 1 - u^2:
(dy / dx) = (1 / (1 + (1 - x)^2)) * (dx / (2 * sqrt(x) * sqrt(1 - x^2)))

So, the final derivative of y = arctan(arcsin(sqrt(x))) is:
(dy / dx) = dx / (2 * sqrt(x) * sqrt(1 - x^2) * (2 - 2x))