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anonymous
 5 years ago
differentiate y=x/square root (x^2+1)
anonymous
 5 years ago
differentiate y=x/square root (x^2+1)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Use the quotient rule. Do you know what that is?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes i just wanted to know the answer because i keep getting something different to the answer given

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so i'm not sure where i went wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think you forgot x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and the correct answer according to the solution is 1/(x^2+1)^(3/2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got \[(1/(x^2+1) )x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The derivative is: \[\frac{x^2}{\left(1+x^2\right)^{3/2}}+\frac{1}{\sqrt{1+x^2}}=\frac{1}{\left(1+x^2\right)^{3/2}} \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0robtobey could you show me your working out before reaching what you had come up with in the above post thanks!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Used Mathematica 8. The request for the derivative from that program was:\[D\left[\frac{x}{\sqrt{1+x^2}},x\right] \] Browse over to WolframAlpha.com and enter: derivative of x/square root (x^2+1) Seclect show steps See the attachment

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Product rule: \[ \dfrac{\mathbb{d}}{\mathbb{d}x}(u\cdot v)=u\cdot \dfrac{\mathbb{d}v}{\mathbb{d}x}+v\cdot \dfrac{\mathbb{d}u}{\mathbb{d}x} \] where u, v are functions of x. The quotient rule is just extra baggage.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you robtobey,romero and INewton!! ^.^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0¬_¬ It was all robtobey.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but INewton made a lovely conclusion to the question!
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