hpfan101
  • hpfan101
Differentiate: \[\frac{ x^2e^x }{ x^2+e^x }\]
Calculus1
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chestercat
  • chestercat
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Empty
  • Empty
Are you familiar with the quotient rule?
UnkleRhaukus
  • UnkleRhaukus
\[\large\left(\frac uv\right)' = \frac{u'v-uv'}{v^2}\]
UnkleRhaukus
  • UnkleRhaukus
@hpfan101, can you identify \(u\) and \(v\)? can you find the derivatives \(u'\) and \(v'\)?

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hpfan101
  • hpfan101
I'm somewhat familiar with the quotient rule.
hpfan101
  • hpfan101
Would it be: \[\frac{ (x^2e^x)'(x^2+e^x)-(x^2e^x)(x^2+e^x)' }{ (x^2+e^x)^2 }\]
UnkleRhaukus
  • UnkleRhaukus
yeah, that's right, now simplify those derivatives
hpfan101
  • hpfan101
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 \[2xe^x\]?
UnkleRhaukus
  • UnkleRhaukus
use the product rule \[(fg)' = f'g+fg'\]
hpfan101
  • hpfan101
Oh okay. So it would be: \[(x^2)'(e^x)+(x^2)(e^x)'= 2x(e^x)+x^2(e^x) =2xe^x+x^2e^x\]
UnkleRhaukus
  • UnkleRhaukus
that's right and if you factor it \(= (2x+x^2)e^x\)
hpfan101
  • hpfan101
Oh okay! Thank you! I think I can figure out the rest.
UnkleRhaukus
  • UnkleRhaukus
to compute\[(x^2+e^x)'\] use the sum rule \[(a+b)'=a'+b'\]
hpfan101
  • hpfan101
Okay, will do! Thanks!
UnkleRhaukus
  • UnkleRhaukus
NB: expanding the square in the denominator will not simplify your result (in this case)
hpfan101
  • hpfan101
Ah, okay. I see. It'd be easier to leave the denominator how it is.

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