Find an equation of the line tangent to teh curve at the given point.

y=x^2 * e^-x , (1, 1/e)

Question

Find an equation of the line tangent to teh curve at the given point.

y=x^2 * e^-x  , (1, 1/e)

anonymous · Answer

take the derivative using the product rule

anonymous · Answer

you get 
$$y'=2xe^{-x}-x^2e^{-x}$$ and then replace x by 1 to get the slope

anonymous · Answer

$$f'(1)=\frac{2}{e}-\frac{1}{e}=\frac{1}{e}$$
then use point slope formula

anonymous · Answer

im confused about how u got 2/e

kira_yamato · Answer

Just plug in x = 1 into the f'(x) or y'

anonymous · Answer

i got 
$$\frac{2}{e}$$ because the first term of the derivative is 
$$2xe^{-x}$$ and if i replace x by 1 i get 
$$2\times 1\times e^{-1}=2e^{-1}=\frac{2}{e}$$

anonymous · Answer

so is that the slope (2/e) that i use 2 plug into teh slope-interept equation?
 y - (1/e) = 2/e (x- 1)

anonymous · Answer

y = 2/e(x) - 1/e

anonymous · Answer

no the slope is not 
$$\frac{2}{e}$$ it is 
$$\frac{1}{e}$$

anonymous · Answer

i thought wen we evaluate teh derivative at the x value that wud b the slope...how is 1/e the slope

anonymous · Answer

plz explain..i know im slow at math