Question

Differentiate and evaluate at x=0, y=(x^2-2x+9) e^x?

Answer 1

the derivatice should be \[\huge (x^2-2x+9)*e^x+(2x-2)\]

Answer 2

or

Answer 3

\[\huge (x^2+7)*e^x\]

Answer 4

I thought it should be \[e^x (x^2-2x+9)+e^x (2x-2)=e^x(x^2-7)\]

Answer 5

if x=0 then it will be

Answer 6

You don't do anything with the e^x?

Answer 7

Yeah, you don't.

Answer 8

\[\huge 7e^x\]

Answer 9

You can try that out from first principle.

Answer 10

apply the product rule first

Answer 11

Whats the product rule?

Answer 12

I can't find it

Answer 13

\[\frac{d}{dx}f(x)g(x)=f'(x)g(x)+f(x)g'(x)\]

Answer 14

same as quotient but without the denominator?

Answer 15

d(uv)=u(dv)+v(du)

Answer 16

Yeah, you can derive the quotient rule from product rule.

Answer 17

They are the same literally.

Answer 18

where did you guys get (2x-2)?

Answer 19

\[\frac{d}{dx}(x^2-2x+9)=2x-2\]

Answer 20

yup!

Answer 21

Oh wow, lol hahah I see now, thanks

Answer 22

Does it matter where the e^x is? like in front of the parenthesis or behind