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## emcrazy14 2 years ago Differentiate; x^2/(1+x)^2 Delete Cancel Submit

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1. ajprincess
• 2 years ago
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This is of the form u/v. (u/v)'=(u'v-v'u)/v^2 Here u=x^2 and v=(1+x)^2. (x^n)'=nx^(n-1) Can u do it @emcrazy14?

2. emcrazy14
• 2 years ago
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I got the answer -2x/(1+x)^3. Is it correct?

3. experimentX
• 2 years ago
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differentiate: 1/(1/x + 1)^2

4. ajprincess
• 2 years ago
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check the sign @emcrazy14. It shud be 2x/(1+x)^3

5. emcrazy14
• 2 years ago
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Ok, thanks. Do I multiply (x+1) with a negative sign to change it into (1+x) so that it cancels with the the denominator (1+x)^4?

6. experimentX
• 2 years ago
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just ask the wolf for differentiation Question.

7. ajprincess
• 2 years ago
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nope. (x+1)=(1+x)

8. emcrazy14
• 2 years ago
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Then how oes it become +2x? :O

9. emcrazy14
• 2 years ago
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@experimentX : wolf? :O *does

10. experimentX
• 2 years ago
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11. ajprincess
• 2 years ago
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|dw:1352469822062:dw||dw:1352469940166:dw||dw:1352470079425:dw|

12. experimentX
• 2 years ago
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$x^2/(1+x)^2 = 1/(1/x + 1)^2 = ({1 \over x} + 1)^{-2}$ use chain rule ... this should be shorter.

13. emcrazy14
• 2 years ago
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@ajprincess : Get it now. Thank you so much! :D

14. ajprincess
• 2 years ago
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welcome:) @experimentX's method is much more simpler. U can use that too.:)

15. emcrazy14
• 2 years ago
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@experimentX : Once I perfect this one, I'll use the one you mentioned. Thank you! :)

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