0
$\begingroup$

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?

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

$\endgroup$
  • $\begingroup$ Just use $\bigl(\sqrt x\bigr)'$ where required. $\endgroup$ – Bernard Jul 4 '18 at 8:11
1
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.