I have a function that I want to find the derivative of using the difference quotient definition of a derivative. The function is:


therefore, using the difference quotient definition, we have:


this is equal to:


I tried using (a+b)(a-b) = a^2 - b^2 to get rid of the square roots in the numerator, but the denominator gets pretty huge, so I'm not sure if this is a wise path to proceed down.

At this point I get stuck with the algebra. I understand that this is a simple derivative to take using the quotient rule, but I'm trying to practice taking limits, and to learn useful algebra tricks.

A step by step computation would be helpful. Many thanks


You properly wrote$$\lim_{h\to0}\frac{1}{h}\left(\frac{(x+1)\sqrt{x+h}-(x+h+1)\sqrt{x}}{(x+1)(x+h+1)}\right)$$ Now, it is clear that the problem is within $\sqrt{x+h}$. But, you could use the generalized binomial theorem to write $$\sqrt{x+h}=\sqrt{x}+\frac{h}{2 \sqrt{x}}+\cdots$$ So, for the numerator $$A=(x+1)\sqrt{x+h}-(x+h+1)\sqrt{x}=(x+1)\left(\sqrt{x}+\frac{h}{2 \sqrt{x}}+\cdots \right)-(x+h+1)\sqrt{x}$$ Expanding and simplifying $$A=\frac{h}{2 \sqrt{x}}-\frac{h \sqrt{x}}{2}=h\frac{1- x}{2 \sqrt{x}}$$ and the overall problem becomes simple.

  • $\begingroup$ Thank you for this answer, the handling roots with the generalised binomial theorem is certainly a new trick for me. However, I do not see how the second equation for A follows from the binomial expansion in the first. I can see a simplification is possible in the first equation (which removes just the x^0.5 term from the binomial expansion) but I can't see any further simplification (I don't have much experience dealing with infinite series). I would be most grateful if you could elucidate. $\endgroup$ – Jinglesting Jul 6 '17 at 10:41
  • $\begingroup$ @Jinglesting. You do not care about the next terms of the binomial expansion since they are $h^2$, $h^3$ and so on. This is why I wrote $+\cdots$. Is it more clear now ? $\endgroup$ – Claude Leibovici Jul 6 '17 at 10:55
  • $\begingroup$ Yes I understand now, thank you for explaining. I've sometimes seen things like o(h). Could o(h^2) be used here? Thanks $\endgroup$ – Jinglesting Jul 6 '17 at 11:05
  • $\begingroup$ @Jinglesting. Yes, for sure. In the same spirit, you could use Taylor series to $O(h^2)$ writing $$\sqrt{x+h}=\sqrt{x} \sqrt{1+\frac hx}=\sqrt{x} \left(1+\frac h {2x}+O(h^2)\right)$$ $\endgroup$ – Claude Leibovici Jul 6 '17 at 11:20

Alternatively, you can use the definition

$$f'(a)=\lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a}.$$

This gives

$$\frac{f(x)-f(a)}{x-a}=\frac{\frac{\sqrt x}{x+1}-\frac{\sqrt a}{a+1}}{x-a}=\frac{(a+1)\sqrt x-(x+1)\sqrt a}{(a+1)(x+1)(x-a)}$$

$$=\frac{(a\sqrt x-x\sqrt a)+(\sqrt x-\sqrt a)}{(a+1)(x+1)(x-a)}=\frac{-\sqrt {ax}\cdot (\sqrt x-\sqrt a)+(\sqrt x-\sqrt a)}{(a+1)(x+1)(x-a)}$$

$$=\frac{1-\sqrt {ax}}{(a+1)(x+1)}\cdot\frac{\sqrt x-\sqrt a}{x-a}$$

As $x \rightarrow a$, the first factor becomes


and the second

$$\frac{\sqrt x -\sqrt a}{x-a}=\frac{x-a}{(x-a)(\sqrt x +\sqrt a)}=\frac{1}{\sqrt x+\sqrt a}\rightarrow \frac{1}{2\sqrt a}$$

  • $\begingroup$ Good to know another trick, I did not realise there existed an alternative, equivalent definition. Thank you $\endgroup$ – Jinglesting Jul 6 '17 at 10:44

Using your idea, just ignore the denominator until later. If you go through the numerator, you will see that $$\begin{align*} &\left((x+1)\sqrt{x+h}-(x+h+1)\sqrt{x}\right)\cdot\left((x+1)\sqrt{x+h}+(x+h+1)\sqrt{x}\right)\\ =&(x+h)(x+1)^2-x(x+h+1)^2\\ =&hx^2+2hx+h+x^3+2x^2+x-h^2x-2hx^2-2hx-x^3-2x^2-x\\ =&-hx^2-h^2x+h\\ =&-h(x^2+xh-1) \end{align*}$$ at which point you notice this factor of $h$ will get along nicely with the factor of $\frac{1}{h}$...

The limit then becomes $$f'(x)=\lim_{h\to 0}-\frac{x^2+xh-1}{(x+1)(x+h+1)\left((x+1)\sqrt{x+h}+(x+h+1)\sqrt{x}\right)}$$

  • $\begingroup$ So it turns out you can just take the limit here after all then, I wasn't sure. Limits are a pretty recent thing to me. Thanks for your answer. $\endgroup$ – Jinglesting Jul 6 '17 at 10:43

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