Derivative with a Square root in Denominator
$f(x) = \dfrac{-3}{\sqrt{3x^2 + 3}}$
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.
$\endgroup$ 14 Answers
$\begingroup$One route to go is to write your function $$f(x) = -3(3x^2 + 3)^{-1/2}$$ and then use the chain rule!
$$f'(x) = -(1/2)(-3) (3x^2 + 3)^{-3/2}\cdot(6x)= 9x(3x^2 + 3)^{-3/2} $$
$$= \dfrac{9x}{(3x^2 + 3)^{3/2}}$$
$\endgroup$ 2 $\begingroup$Exponent rules! Remember that $\dfrac{1}{\sqrt{x}} = x^{-1/2}$. Then use the power and chain rule.
$\endgroup$ $\begingroup$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,
$$ f'(x) = \frac{-1}{2} \frac{-3}{(3x^2 + 3)^\frac{3}{2}}(6x) = \frac{9x}{(3x^2 + 3)^\frac{3}{2}}$$
$\endgroup$ $\begingroup$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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