$f(x) = \dfrac{-3}{\sqrt{3x^2 + 3}}$
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.
One route to go is to write your function $$f(x) = -3(3x^2 + 3)^{-1/2}$$ and then use the chain rule!
$$f'(x) = -(1/2)(-3) (3x^2 + 3)^{-3/2}\cdot(6x)= 9x(3x^2 + 3)^{-3/2} $$
$$= \dfrac{9x}{(3x^2 + 3)^{3/2}}$$
Exponent rules! Remember that $\dfrac{1}{\sqrt{x}} = x^{-1/2}$. Then use the power and chain rule.
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,
$$ f'(x) = \frac{-1}{2} \frac{-3}{(3x^2 + 3)^\frac{3}{2}}(6x) = \frac{9x}{(3x^2 + 3)^\frac{3}{2}}$$
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?
