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$g(x)=\frac{x^{1/4}}{x^3+1}$, find $g'(x)$. Use the quotient rule.

My attempt was:

\begin{align}g'(x)&=\frac{\frac{1}{4}(x^{-3/4})(x^3+1)-(3x^2)(x^{1/4})}{(x^3+1)^2}\\ g'(x)&=\frac{(x^3+1)-(3x^2)(x^{1/4})}{4(x^{3/4})(x^3+1)^2}\end{align}

I am $100\ \%$ sure of the denominator which is the easy part, but or the numerator I am stuck.

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  • $\begingroup$ In the second step, when you moved the "4" and the factor $ \ x^{-3/4} \ $ to the denominator, you forgot to also take factors of $ \ \frac{1}{4} \ $ and $ \ x^{-3/4} \ $ from the second term in the numerator. That term should become $ \ 4 \ (3x^2) \ (x^1) \ $ or $ \ 12x^3 \ $ ... $\endgroup$ Jun 19, 2014 at 2:35

2 Answers 2

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\begin{align} \frac{d}{dx}\frac{x^{\frac 14}}{x^3+1} &=\frac{\frac{d}{dx}(x^{\frac 14})(x^3+1)+(x^{\frac 14})\frac{d}{dx} (x^3+1)}{(x^3+1)^2} \\ &=\frac{\frac 14x^{-\frac 34}(x^3+1)+x^{\frac 14}(3x^2)}{(x^3+1)^2} \end{align}

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$g'(x)=\frac{\frac{1}{4}(x^{-3/4})(x^3+1)-(3x^2)(x^{1/4})}{(x^3+1)^2}$ this step is correct. But in next step you should multiply both the terms in numerator by $4x^{\frac{3}{4}}$ I think, the numerator will be $-11x^3+1$.

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  • $\begingroup$ why downvote? I just corrected the calculation, is there something wrong in my answer? $\endgroup$
    – puru
    Jun 25, 2014 at 6:52

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