Recall that the derivative of the natural exponential function is itself. Also, it's all about the chain rule:
Hi, I'm having a little trouble of finding the derivative of something like this:
y=e^{-2x + (x^2)}
So I'm a bit confused.
Like my teacher shows me the derivatove of 4e^{sinx} = 4e^{sinx} * cosx
There, I tried the same method, doing y' = e^{2x + (x^2)} * e^{-2x + 2x }Though... in the back of the book, it had 2(x-1)e^{-2x + (x^2) }
I searched up this up, and found that there are rules like this?
But after that... I end up with 2 'e'
I'm not really sure right now.
Can someone explain?
Thanks
Use the chain rule. The outer function is e^(that messy exponent). The inner function is the messy exponent, the -2x+(x^2). So you want derivative of the outer (with the messy exponent still in place) times derivative of the inner. Derivative of the outer: piece of cake, e is its own derivative, right? That's one reason mathematicians are so enamored of e. Derivative of the inner: I'm betting you're good at using the power rule by this point, right? You should wind up with 2(x-1) after factoring. If you look at the answer from your book, you'll see they've put deriv of the inner first, but there it is: deriv of the messy exponent times e^(that stuff).
Yeah, you got it. Of course, if the outer function hadn't been e (which is its own derivative), then you would have needed whatever the derivative of the outer function turned out to be, multiplied by the derivative of the inner. Good luck to you!