How to find a derivative using the quotient rule with a square root in the denominator?

Question

I am trying to solve the assignment below with the quotient rule. Every time I get stuck on a similar assignment which has a square root inside the numerator of a fraction. What do I do with the square root in the numerator and how do I find the derivative?

$$\frac{\sqrt x+3}{x}$$

Answer 1

$\frac{d}{dx}\frac{\sqrt{x}+3}{x}=$

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

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

because

1. $\frac{x}{\sqrt{x}}=x^{1-\frac{1}{2}}=\sqrt{x}$

2. $\frac{d}{dx}\frac{f}{g}=\frac{f’g-g’f}{g^2}$ and $\frac{d}{dx}x^\alpha=\alpha x^{\alpha-1}$ for every $\alpha\in \mathbb{R}$

3. $\frac{d}{dx}\sqrt{x}= \frac{d}{dx} x^{\frac{1}{2}}=\frac{1}{2}x^{\frac{1}{2}-1}=\frac{1}{2\sqrt{x}}$

Answer 2

The fraction rule says that $$ \left(\frac{\sqrt x + 3}{x}\right)' = \frac{(\sqrt x + 3)'\cdot x - (\sqrt x + 3)\cdot (x)'}{x^2} $$ Now we need to find the different derivatives in the numerator. The second one is easy: $(x)' = 1$. For the first derivative, $(\sqrt x + 3)'$, you use several rules. First differentiation of sum: $$ (\sqrt x + 3)' = (\sqrt x)' + (3)' $$ Then, separately, differentiation of square root, and differentiation of a constant: $$ (\sqrt x)' + (3)' = \frac1{2\sqrt x} + 0 $$ This we now insert into our original fraction: $$ \frac{(\sqrt x + 3)'\cdot x - (\sqrt x + 3)\cdot (x)'}{x^2} = \frac{\frac{1}{2\sqrt x}\cdot x - (\sqrt x + 3)\cdot 1}{x^2} $$ and with that we're done with the differentiation. The rest is algebraic simplification, and then you're finished.