## hpfan101 one year ago Differentiate: $\frac{ x^2e^x }{ x^2+e^x }$

1. Empty

Are you familiar with the quotient rule?

2. UnkleRhaukus

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

3. UnkleRhaukus

@hpfan101, can you identify $$u$$ and $$v$$? can you find the derivatives $$u'$$ and $$v'$$?

4. hpfan101

I'm somewhat familiar with the quotient rule.

5. hpfan101

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

6. UnkleRhaukus

yeah, that's right, now simplify those derivatives

7. 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 $2xe^x$?

8. UnkleRhaukus

use the product rule $(fg)' = f'g+fg'$

9. hpfan101

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

10. UnkleRhaukus

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

11. hpfan101

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

12. UnkleRhaukus

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

13. hpfan101

Okay, will do! Thanks!

14. UnkleRhaukus

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

15. hpfan101

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