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