Question

Given f(x)=[[3e^(1-(x^2))]*ln(x)] find the equation of the tangent line at x=1

Answer 1

you know how to derivative?

Answer 2

\[\frac{d}{dx}\left(\left(3 e^{1-x^2}\right) \ln (x)\right)=-\frac{3 e^{1-x^2} \left(2 x^2 \ln (x)-1\right)}{x}\]
\[-\frac{3 e^{1-x^2} \left(2 x^2 \ln (x)-1\right)}{x}~,~x=1\]

\[\frac{d}{dx}=3\]

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y-y1=m(x-1)

Answer 3

m=3

Answer 4

use y-y1=m(x-1)

Answer 5

Is the answer y=3x-3?

Answer 6

yes