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?

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

  • $\begingroup$ Just use $\bigl(\sqrt x\bigr)'$ where required. $\endgroup$ – Bernard Jul 4 '18 at 8:11


$\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} $



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

  • $\begingroup$ On the second and third step, how did you merge the square root from both terms? $\endgroup$ – tomwassing Jul 4 '18 at 8:16
  • $\begingroup$ what do you think? Other question? $\endgroup$ – Federico Fallucca Jul 4 '18 at 8:18
  • $\begingroup$ Hmm, my weakness lies in the simplification. I can't see how I can simplify: 1/2*sqrt(x) - sqrt(x) - 3 $\endgroup$ – tomwassing Jul 4 '18 at 8:25
  • $\begingroup$ Now I think it is more clear $\endgroup$ – Federico Fallucca Jul 4 '18 at 8:33
  • $\begingroup$ Thank you! It is now. I didn't see that I could subtract from the 1/2. I thought it was not allowed to subtract from a product. For example = 6*5-3, I can't just subtract 3 from 6 or 5. But I see now how it is possible in this situation. Many thanks to you! $\endgroup$ – tomwassing Jul 4 '18 at 8:38

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.


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