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$f(x) = \dfrac{-3}{\sqrt{3x^2 + 3}}$

I can't seem to figure this problem out. I think you would make the bottom(3x^2+3)^(1/2) and then use the chain rule on bottom and then use the quotient rule. This is the only question I cant seem to figure out on my homework so if you could give step by step detailed instructions i'd be forever grateful. Thanks for your time.

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  • $\begingroup$ You could do that. But the Quotient rule is what you'd use first. $\endgroup$ – David Mitra Feb 25 '14 at 17:47
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One route to go is to write your function $$f(x) = -3(3x^2 + 3)^{-1/2}$$ and then use the chain rule!

$$f'(x) = -(1/2)(-3) (3x^2 + 3)^{-3/2}\cdot(6x)= 9x(3x^2 + 3)^{-3/2} $$

$$= \dfrac{9x}{(3x^2 + 3)^{3/2}}$$

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  • $\begingroup$ Have a wonderful time my friend. I'll try to get back soon. One of my article is going to be published at Korean Mathematical Society. :-) $\endgroup$ – mrs Feb 26 '14 at 9:27
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    $\begingroup$ I'm so glad to hear from you, AND so glad that you're going to be published! $\endgroup$ – Namaste Feb 26 '14 at 13:36
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Exponent rules! Remember that $\dfrac{1}{\sqrt{x}} = x^{-1/2}$. Then use the power and chain rule.

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Just use the chain rule, first bringing up

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

Then, taking the derivative of what we have raised to the (-1/2) power is just the use of the chain rule, and we will have,

$$ f'(x) = \frac{-1}{2} \frac{-3}{(3x^2 + 3)^\frac{3}{2}}(6x) = \frac{9x}{(3x^2 + 3)^\frac{3}{2}}$$

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An alternative approach: Quotient Rule tells you that $$f'(x)=\cfrac{\sqrt{3x^2+3}\cdot\frac{d}{dx}[-3]-(-3)\cdot\frac{d}{dx}\left[\sqrt{3x^2+3}\right]}{\left(\sqrt{3x^2+3}\right)^2}.$$

Since $\frac{d}{du}\left[\sqrt u\right]=\frac1{2\sqrt u},$ then by Chain Rule, $$\frac{d}{dx}\left[\sqrt{3x^2+3}\right]=\frac1{2\sqrt{3x^2+3}}\cdot\frac{d}{dx}[3x^2+3].$$

Can you take it from there?

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