Equation of Tangent Line to y=(lnx)^cosx at (pi/2,1)

[dvaughn]

Find the equation of the tangent line to the curve y=(lnx)^cosx at the point (pi/2, 1)?

The Attempt at a Solution

lny = cosx(ln(ln(x))) d/dx
= -sinx(ln(ln(x))) + cosx/(ln(x)(x))
y' = y(-sinx(ln(ln(x))) + cosx/(ln(x)(x)))
this is the part where I get stuck

 

[jhae2.718]

Find the equation of the tangent line to the curve y=(lnx)^cosx at the point (pi/2, 1)?

The Attempt at a Solution

lny = cosx(ln(ln(x))) d/dx
= -sinx(ln(ln(x))) + cosx/(ln(x)(x))
y' = y(-sinx(ln(ln(x))) + cosx/(ln(x)(x)))
this is the part where I get stuck

When you get to
y'=y\left[ {\frac{\cos\left(x\right)}{x\ln\left(x\right)}-\sin\left(x\right)\ln\left(\ln\left(x\right)\right)}\right]
substitute \ln\left(x\right)^{\cos\left(x\right)} for y.

Then take y'\left(\frac{\pi}{2}\right) as the slope of your tangent line and find what y-intercept will put the line through \left(\frac{\pi}{2},1\right).