Question

Differentiate:
\[\frac{ x^2e^x }{ x^2+e^x }\]

Answer 1

Are you familiar with the quotient rule?

Answer 2

\[\large\left(\frac uv\right)' = \frac{u'v-uv'}{v^2}\]

Answer 3

@hpfan101, can you identify \(u\) and \(v\)?
can you find the derivatives \(u'\) and \(v'\)?

Answer 4

I'm somewhat familiar with the quotient rule.

Answer 5

Would it be: \[\frac{ (x^2e^x)'(x^2+e^x)-(x^2e^x)(x^2+e^x)' }{ (x^2+e^x)^2 }\]

Answer 6

yeah, that's right,
now simplify those derivatives

Answer 7

Okay, what I'm not sure about is the first set we would find the derivative of. Would the derivative of \[(x^2e^x)'\] be \[2xe^x\]?

Answer 8

use the product rule
\[(fg)' = f'g+fg'\]

Answer 9

Oh okay.
So it would be: \[(x^2)'(e^x)+(x^2)(e^x)'= 2x(e^x)+x^2(e^x) =2xe^x+x^2e^x\]

Answer 10

that's right
and if you factor it \(= (2x+x^2)e^x\)

Answer 11

Oh okay! Thank you! I think I can figure out the rest.

Answer 12

to compute\[(x^2+e^x)'\]
use the sum rule
\[(a+b)'=a'+b'\]

Answer 13

Okay, will do! Thanks!

Answer 14

NB: expanding the square in the denominator will not simplify your result (in this case)

Answer 15

Ah, okay. I see. It'd be easier to leave the denominator how it is.