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calculus

posted by on .

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?

  • calculus - ,

    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.

  • calculus - ,

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

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