How to find the derivative of y= x^-1/e^-x

[intenzxboi]

Homework Statement

y = ((x)^−1) (e^−x)

The Attempt at a Solution

Solving this out i got
[(x^-1) - (x^-2)] e^-x

however the book is telling me the answer is
−(x^−1 + x^−2)e^−x

wouldnt that give me (-x^−1 - x^−2)e^−x ??
what did i do wrong??

 

[Mark44]

Homework Statement

y = ((x)^−1) (e^−x)

The Attempt at a Solution

Solving this out i got
[(x^-1) - (x^-2)] e^-x

however the book is telling me the answer is
−(x^−1 + x^−2)e^−x

wouldnt that give me (-x^−1 - x^−2)e^−x ??
what did i do wrong??

Apparently you lost a sign somewhere. I'm guessing that you forgot the factor of (-1) when you differentiated e^(-x).
$$d/dx(x^{-1}e^{-x}) = x^{-1} e^{-x} (-1) - x^{-2}e^{-x}$$
$$= -x^{-1}e^{-x} - x^{-2} e^{-x}$$
which agrees with the book's answer.

BTW, you're not "solving this out;" you're differentiating this function or calculating the derivative of the given function.

 

[intenzxboi]

Thanks i thought the derivative of e^-x stayed the same

 

[Hootenanny]

Thanks i thought the derivative of e^-x stayed the same

$$\frac{d}{dx}e^x = e^x$$

However, in general (using the chain rule)

$$\frac{d}{dx} e^{f\left(x\right)} = f^\prime\left(x\right)e^{f\left(x\right)}$$

In your case

$$f\left(x\right) = -x$$

So

$$\frac{d}{dx}e^{-x} = \frac{d}{dx}\left(-x\right)e^{-x} = - e^{-x}$$

Do you follow?