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January 30, 2015

January 30, 2015

Posted by **Chelsea** on Thursday, March 10, 2011 at 6:53pm.

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?

- calculus -
**bobpursley**, Thursday, March 10, 2011 at 7:02pmln 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.

- calculus -
**Chelsea**, Thursday, March 10, 2011 at 8:10pmI'm sorry, but I'm confused on the 2nd part?

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