Differentiate expression involving reciprocal of square roots. I need to differentiate $$5\over 2+\sqrt{1+3x}$$
I can get the answer from Wolfram Alpha but I'm trying to understand the working. Do I use the chain rule? My calculus is at the basic level.
 A: Rewrite the function as $$f(x) = 5(2+\sqrt{1+3x})^{-1}$$ now apply the chain and power rule to get $$f'(x) = 5\cdot -1 \cdot (2+\sqrt{1+3x})^{-2} \cdot \frac{\mathrm{d}}{\mathrm{d}x} (2+\sqrt{1+3x})$$
This yields $$f'(x) = -\frac{5}{(2+\sqrt{1+3x})^2} \cdot \frac{3}{2\sqrt{1+3x}}$$
You can re-write this as $$f'(x) = -\frac{15}{2\sqrt{1+3x}\cdot \left(2+\sqrt{1+3x}\right)^2}$$

This is just an application of the fact that $$\frac{\mathrm{d}}{\mathrm{d}x}(y^a) = a y^{a-1} \cdot y'$$
A: Yes you can use from Chain rule bout it is not necessary.
$$(f(g(x))^\prime=g^\prime(x)f^\prime(g(x)) \tag{1}$$
if
$$f(x)=\dfrac{5}{2+x} ‎\longrightarrow‎f^\prime(x)=\dfrac{-5}{(2+x)^{2}}\tag{2.1}$$
$$g(x)=\sqrt {1+3x}‎\longrightarrow‎g^\prime(x)=\dfrac{3}{2\sqrt{1+3x}}\tag{2.2}$$
From $(1),(2.1) $ and $(2.2)$:
$$(f(g(x))^\prime=\dfrac{3}{2\sqrt{1+3x}}\cdot\dfrac{-5}{(2+\sqrt{1+3x})^{2}}$$
And in this way to get you the answer.
A: Just use the Quotient Rule :
$\frac{d}{dx}\frac{f}{g} = \frac{f'g-g'f}{g^2}$ . In this case set $f = 5$ and $g =  2+\sqrt{1+3x}$ .  Hence :
$f'g = 0$ 
$g'f =\frac{15}{2\sqrt{1+3x}}$
$g^2 = (2+\sqrt{1+3x})^2 $
Combining all of the above we get :

$$\frac{d}{dx}\frac{f}{g} = \frac{f'g-g'f}{g^2} = -\frac{15}{2\sqrt{1+3x}\cdot \left(2+\sqrt{1+3x}\right)^2}$$

A: The chain rule is as follows:
If we define $h(x) = (f \circ g)(x)$:
$$h'(x) = f'(g(x)) \cdot g'(x)$$
There are several ways to compute the derivative of this expression. One way is to use the quotient rule, however because the numerator is just a number this does not seem like the best option. A good way to compute such a derivative is to rewrite it as:
$$5(2+\sqrt{1+3x})^{-1}$$
Now let us use the chain rule (note that we can 'ignore' the 5 for now and just multiply it back on later because it is just a number. 
$$h'(x) = 5(-(2+\sqrt{1+3x}))^{-2} \cdot (2+\sqrt{1+3x})'$$
$$ = -5(2+\sqrt{1+3x})^{-2} \cdot (2+\sqrt{1+3x})'$$
Now we must take the derivative of $2+\sqrt{1+3x}$.
$$(2+\sqrt{1+3x})'$$
$$ = (2+{(1+3x)}^{1/2})$$
$$=\frac{1}{2}(1+3x)^{-1/2} \cdot 3$$
$$=\frac{3}{2 \sqrt{1+3x}}$$
The 3 comes from the derivative of $1+3x$. Going back to $h'(x)$ and subbing in the values that we know:
$$h'(x) = -5(2+\sqrt{1+3x}))^{-2} \cdot \frac{3}{2 \sqrt{1+3x}}$$
$$=-\frac{15}{(2+\sqrt{1+3x})^{2} \cdot (2 \sqrt{1+3x})}$$
$$ = \boxed{-\frac{15}{2 \sqrt{1+3x}\cdot (2+\sqrt{1+3x})^{2}}}$$
A: Use
$$
(y^a)'=ay^{a-1} y'
$$
for $a=-1$ and then for $a=1/2$.
