Asked by Anonymous
What is the slope of the tangent for the function h(x)=sin(e^-lnx) at x=1/pi
Answers
Answered by
oobleck
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
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
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