Question

What is the slope of the curve y=x*e^x at x=-2?

Answer 1

Friend of mine said that you take dy/dx and that is e^x + xe^x but I don't see how. Obviously you would just plug in x=-2 after that. Let me rewrite with the euqation feature here: \[dy/dx = e ^{x}+xe ^{x} ???\]

Answer 2

I'm assuming you are looking for the slope of the tangent line at x=2?
If so, then you would take the derivative or y' of x*e^x, which would be y'= x*e^x+e^x. Then using the derivative, you find the slope by doing y'(-2). Substitute -2 into the derivative equation for x. So you would have y'(-2) = (-2)*e^(-2) + e^(-2). I would leave the slope in terms of e^x unless your were told to have a decimal. So the answer would be y'(-2) or m= -2e^(-2)+e^(-2). Simplify and you should get m= -e^(-2) in terms of e^x or as a decimal = -0.135

Answer 3

Sorry, in my first sentence, that should be at x= -2

Answer 4

Are you familiar with the product rule for taking derivatives?

Answer 5

Thank you Raquella! It seems my friend must be right. I just don't quite see how that is the derivative. And yes Jason, I thought I was! But clearly I was mistaken :(

Answer 6

You need to take derivative of the product.
y' = F'S + S'F

Answer 7

NP!! For the derivative of x*e^x, you have to apply the product rule, but keep in mind that the derivative of e^x is e^x*derivative of x if its a greater power than one.
For example if you had e^(xsquared) the derivative would be e^(xsquared)*2x. Hope that helps.

Answer 8

Clears it all up for me guys, thanks so much! I don't know why my mind blanked on how to properly do the product rule, I guess the variables scared me XD

Answer 9

maybe youre tired :D

Answer 10

:D Goodluck!