How should I take the derivative of the following:
$$\frac{t^2}{(1+t^4)^{1/2}}$$
I know the answer and I have tried quotient rule and product rule and I can't seem to succeed.
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1Are you sure your answer isn't equivalent to the answer you are looking for? – John Habert Feb 11 '14 at 06:07
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Yes, I am sure. So what strategy would you use? – user127384 Feb 11 '14 at 06:14
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Did you remember to do the chain rule when differentiating $(1+t^4)^{1/2}$? – mlg4080 Feb 11 '14 at 06:16
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Quotient or product will give the correct answer. But based on the functions involved, the answer you want may have been simplified in a way that looks different from what you get from just using the rules. – John Habert Feb 11 '14 at 06:16
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For the product rule I get f(t)=t^2 f'(t)=2t g(t)=(1+t^4)^(-1/2) g'(t)= (-1/2)(1+t^4)^(3/2)4x^3 Then t^2(-1/2)(1+t^4)^(3/2)4x^3+(1+t^4)^(-1/2)2t Is this so far correct? – user127384 Feb 11 '14 at 06:23
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Exponent should be -3/2 and you have an $x$ instead of $t$ but the rest looks fine. – John Habert Feb 11 '14 at 06:38
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You were right then I am unable to simplify this correctly. – user127384 Feb 11 '14 at 06:47
2 Answers
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If you're having trouble with the Quotient Rule you could reorganize the expression $$ \frac{t^2}{(1+t^4)^{1/2}} = t^2\cdot\left(1+t^4\right)^{-1/2} = \left(t^{-4}\right)^{-1/2}\cdot\left(1+t^4\right)^{-1/2} = \left(t^{-4}+1\right)^{-1/2} $$ and use the Chain Rule.
Brian Fitzpatrick
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Using the Quotient Rule, we get
$\begin{align} \dfrac{d}{dt}\left(\dfrac{t^2}{(1+t^4)^{\frac{1}{2}}}\right) &= \dfrac{(1+t^4)^{\frac{1}{2}}\cdot 2t-t^2\cdot\frac{1}{2}(1+t^4)^{-\frac{1}{2}}\cdot4t^3}{1+t^4} \\ &= \dfrac{2t(1+t^4)^{\frac{1}{2}}-2t^5(1+t^4)^{-\frac{1}{2}}}{1+t^4} \\ &= \dfrac{2t(1+t^4)^{-\frac{1}{2}}\left((1+t^4)-t^4\right)}{1+t^4} \\ &= \dfrac{2t}{(1+t^4)^{\frac{3}{2}}} \end{align} $
John Habert
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