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Differentiate $\frac{(x^3-x+4)^{-\frac{1}{2}}}{\sqrt{x+2}}$

So I rewrite it into $\frac{1}{\sqrt{x^3-x+4}\sqrt{x+2}}$.

So what way should I use? Product rule or quotient rule? The given answer is $-\frac{2x^3+3x^2-x+1}{(x+2)^{\frac{3}{2}}(x^3-x+4)^{\frac{3}{2}}}$. Hope someone can explain it i detail. Thanks in advance.

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    The product rule and the quotient rule are equivalent. You can use either, they'll give you the same answer. One may take more steps to simplify, that's all. – 2017-02-09
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    Either will work, though it's probably easier to start with the chain rule $\left(1 / \sqrt{f(x)}\right)' =\cdots$ – 2017-02-09
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    logarithmic differentiation – 2017-02-09

5 Answers 5

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It is better you use quotient rule in this case, not because the question needs it, but because the simplification to the given answer is much easier.

I find it difficult to type in the steps, but basically you've got to square the denominator for quotient rule, and the answer has the cube of the denominator (Hope you get my point). Simplification after the numerator is found will make your denominator similar to that of the answer.

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Multiply terms in denominator as both terms have same power.

So we have,

$$\frac{1}{\sqrt{(x^3 - x + 4)(x+2)}}$$

$$\frac{1}{[(x^3 - x + 4)(x+2)]^\frac 12}$$

$$[(x^3 - x + 4)(x+2)]^{\frac {-1}2}$$

Now its derivative,

$$\frac{-1}{2} [(x^3 - x + 4)(x+2)]^{\frac {-3}2} \frac {d}{dx}(x^3 - x + 4)(x+2)$$

Hope you can proceed further.

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I would use the chain rule to start, and then the product rule:

$$\left(\frac{1}{\sqrt{x^3-x+4}\sqrt{x+2}}\right)'$$ $$=\left(\frac{1}{\sqrt{(x^3-x+4)(x+2)}}\right)'$$ $$=\left([(x^3-x+4)(x+2)]^{-\frac{1}{2}}\right)'$$

We apply the chain rule:

$$=-\frac{1}{2}[(x^3-x+4)(x+2)]^{-\frac{3}{2}} \cdot [(x^3-x+4)(x+2)]'$$

and then the product rule:

$$=-\frac{(x^3-x+4)(1)+(3x^2-1)(x+2)}{2(x^3-x+4)^{\frac{3}{2}}(x+2)^{\frac{3}{2}}}$$ $$=-\frac{4x^3+6x^2-2x+2}{2(x^3-x+4)^{\frac{3}{2}}(x+2)^{\frac{3}{2}}}$$ $$=-\frac{2x^3+3x^2-x+1}{(x^3-x+4)^{\frac{3}{2}}(x+2)^{\frac{3}{2}}}$$

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We have $$\frac {d}{dx}[\frac {1}{\sqrt {x+2}\sqrt {x^3-x+4}}]$$ $$= -\frac {\frac {d}{dx} [\sqrt {x+2}\sqrt {x^3-x+4}]}{(\sqrt {x+2}\sqrt {x^3-x+4})^2} $$ $$=-\frac { \frac {d}{dx}[\sqrt {x+2}]\cdot \sqrt {x^3-x+4} + \frac {d}{dx}[\sqrt {x^3-x+4}]\cdot \sqrt {x+2}}{(x+2)(x^3-x+4)} $$ $$=- \frac {\frac {\sqrt {x^3-x+4}}{2\sqrt {x+2}} + \frac {(3x^2-1)\sqrt {x+2}}{2\sqrt {x^3-x+4}}}{(x+2)(x^3-x+4)}$$ $$=-\frac {1}{2 (x+2)^{1.5}\sqrt {x^3-x+4}}- \frac {3x^2-1}{2\sqrt {x+2}(x^3-x+4)^{1.5}} $$ $$=- \frac{2x^3+3x^2-x+1}{(x+2)^{\frac {3}{2}}(x^3-x+4)^{\frac {3}{2}}} $$

Hope it helps.

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Logarithmic differentiation can make life easier $$y=\frac{(x^3-x+4)^{-\frac{1}{2}}}{\sqrt{x+2}}\implies \log(y)=-\frac 12 \log(x^3-x+4)-\frac 12\log(x+2)$$ $$\frac{y'}y=-\frac 12 \frac{3x^2-1}{x^3-x+4}-\frac 12 \frac 1 {x+2}=\frac{-2 x^3-3 x^2+x-1}{(x+2) \left(x^3-x+4\right)}$$ Now, use $y'=y \frac {y'} y$ and simplify.