Question

Help with calculus question! @zepdrix @dan815

Answer 1

sorry, can't calculate that

Answer 2

use chain rule :))

Answer 3

not the product rule?

Answer 4

yeah tht one...... :))
obviously

Answer 5

Is this right?

Answer 6

for first teerm it would be
\[2x e^x + x^2 e^x\]

Answer 7

To save yourself a little bit of work,
you could factor before differentiating,\[\large\rm y=e^x(x^2-x)\]Then you only have to product rule one time, instead of twice.

Answer 8

Oh, ok.

Answer 9

And yea, oops, forgot the x in the first term

Answer 10

i mean the derivative of \[x^2e^x \]

Answer 11

Like this?

Answer 12

woops you did d/dx (x^2-x) in both places! :O

Answer 13

no wiat, ignore that

Answer 14

another try, lol

Answer 15

Cool looks good \c:/

But again, factoring is king!
That's what I would have preferred from this step:\[\large\rm y=e^x(2x-1)+e^x(x^2-x)\]turns into,\[\large\rm y=e^x\left[2x-1+x^2-x\right]\]

Answer 16

Ugh, I just don't think of those things..*sigh*
And that would simplify to \(\Large e^x(x^2+x-1)\)

Answer 17

Thank you so much!

Answer 18

np