# Find equation of tangent line

1. Feb 10, 2010

### dvaughn

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

3. 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

2. Feb 10, 2010

### jhae2.718

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 $\left(\frac{\pi}{2},1\right)$.

Last edited: Feb 10, 2010