Question

What is the slope of the tangent for the function h(x)=sin(e^-lnx) at x=1/pi

Answer 1

for any x, the slope of the tangent line is h'(x)

First, simplify things by noting that e^-lnx = 1/e^lnx = 1/x
h'(x) = cos(1/x) * -1/x^2
h'(1/π) = -π^2 cos(π) = π^2