As of late, I've been trying to use the difference quotient to calculate the derivative of the following function:

$$ f(x) = \frac{x}{\sqrt{x^2+2}} $$

When calculating the limits, the function becomes very complicated. I tried to factorise out the h from the numerator from the expression shown below but I can't straighten out the rest of the equation.

$$ \lim_{h\to 0} \frac{\frac{x+h}{\sqrt{(x+h)^2+2}} - \frac{x}{\sqrt{x^2+2}}}{h} $$

How do I calculate the derivative using this method in an efficient manner? Thank you for your help.


It just requires a bit of algebra. I don't know if there is a slicker way to prove this.

\begin{align} &\frac{x+h}{\sqrt{(x+h)^{2} + 2}} - \frac{x}{\sqrt{x^{2} + 2}} \\ &=\frac{(x+h)\sqrt{x^{2} + 2}-x\sqrt{(x+h)^{2} + 2}}{\sqrt{x^{2}+2}\sqrt{(x+h)^{2} + 2}} \quad \text{cross multiplication} \\ &= \frac{(x+h)\sqrt{x^{2} + 2}-x\sqrt{(x+h)^{2} + 2}}{\sqrt{x^{2}+2}\sqrt{(x+h)^{2} + 2}} \cdot \frac{(x+h)\sqrt{x^{2} + 2}+x\sqrt{(x+h)^{2} + 2}}{(x+h)\sqrt{x^{2} + 2}+x\sqrt{(x+h)^{2} + 2}} \quad \text{conjugate} \\ &=\frac{h(4x + 2h)}{\sqrt{x^{2}+2}\sqrt{(x+h)^{2} + 2}[(x+h)\sqrt{x^{2} + 2}+x\sqrt{(x+h)^{2} + 2}]} \\ &\implies \lim_{h \to 0} \frac{1}{h} \cdot \frac{h(4x + 2h)}{\sqrt{x^{2}+2}\sqrt{(x+h)^{2} + 2}[(x+h)\sqrt{x^{2} + 2}+x\sqrt{(x+h)^{2} + 2}]} \\ &=\lim_{h \to 0} \frac{4x + 2h}{\sqrt{x^{2}+2}\sqrt{(x+h)^{2} + 2}[(x+h)\sqrt{x^{2} + 2}+x\sqrt{(x+h)^{2} + 2}]} \\ &=\frac{2}{(x^{2}+2)^{3/2}} \end{align}


For brevity, let us denote $x_h:= x+h$.

Multiplying/dividing by the conjugate sum to get a difference of squares, the numerator becomes


Then reducing to the common denominator, the numerator is


which tends to $4hx$ while the denominator $(x_h^2+2)(x^2+2)$ tends to $(x^2+2)^2$. Now the conjugate sum was


Putting all together,


  • $\begingroup$ So simpler than mine ! Cheers :-( $\endgroup$ – Claude Leibovici Sep 21 '16 at 7:20
  • $\begingroup$ @ClaudeLeibovici: I made it simple mainly by avoiding a full development. Maybe a cheat :) Cheers. $\endgroup$ – Yves Daoust Sep 21 '16 at 7:24

Let us anticipate a little and use the general product formula $(uv)'=u'v+uv'$, which immediately implies $(u^2)'=2u'u$.

Now we can rewrite the initial identity


and take for granted that $(x^2+2)'=(x^2)'=2x$ which are quite easy to establish.

Then if we differentiate both members,

$$2f'(x)f(x)(x^2+2)+f^2(x)2x=2x$$ and


which leads to the desired result.


$$(u(x)v(x))'=\lim_{h\to0}\frac{u(x+h)v(x)-u(x)v(x+h)}h=\\ \lim_{h\to0}\frac{u(x+h)v(x)-u(x)v(x)+u(x)v(x)-u(x)v(x+h)}h=\\ \lim_{h\to0}\frac{u(x+h)-u(x)}hv(x)+\lim_{h\to0}u(x)\frac{v(x+h)-v(x)}h=\\ u'(x)v(x)+u(x)v'(x).$$

  • $\begingroup$ This one is really elegant ! Thanks for providing this solution. Cheers. $\endgroup$ – Claude Leibovici Sep 21 '16 at 10:38

Mattos gave the elegant way of doing it rigorously.

There is another way I just propose for your curiosity considering $$\frac{x+h}{\sqrt{(x+h)^2+2}}=\frac{x+h}{\sqrt{x^2+2+2hx+h^2}}=\frac{x+h}{\sqrt{x^2+2}\,\sqrt{1+\frac{2hx+h^2}{x^2+2}}}$$ Use the generalized binomial theorem or Taylor expansion or equivalents $$\frac 1 {\sqrt{1+a}}=1-\frac{a}{2}+O\left(a^2\right)$$ So $$\frac 1 {\sqrt{1+\frac{2hx+h^2}{x^2+2}}}\sim 1-\frac{2hx+h^2}{2(x^2+2)}$$ this makes $$\frac{x+h}{\sqrt{(x+h)^2+2}}\sim\frac {x+h}{\sqrt{x^2+2}}\left(1-\frac{2hx+h^2}{2(x^2+2)} \right)$$ $$\frac{x+h}{\sqrt{(x+h)^2+2}}=\frac {x}{\sqrt{x^2+2}}\left(1-\frac{2hx+h^2}{2(x^2+2)} \right)+\frac {h}{\sqrt{x^2+2}}\left(1-\frac{2hx+h^2}{2(x^2+2)} \right)$$ $$\frac{x+h}{\sqrt{(x+h)^2+2}} - \frac{x}{\sqrt{x^2+2}}=-\frac {x}{\sqrt{x^2+2}}\left(\frac{2hx+h^2}{2(x^2+2)} \right)+\frac {h}{\sqrt{x^2+2}}\left(1-\frac{2hx+h^2}{2(x^2+2)} \right)$$ $$\frac{\frac{x+h}{\sqrt{(x+h)^2+2}} - \frac{x}{\sqrt{x^2+2}}}{h}=-\frac {x}{\sqrt{x^2+2}}\left(\frac{2x+h}{2(x^2+2)} \right)+\frac {1}{\sqrt{x^2+2}}\left(1-\frac{2hx+h^2}{2(x^2+2)} \right)$$ Now, making $h \to 0$, the limit is then $$-\frac {x}{\sqrt{x^2+2}}\left(\frac{x}{x^2+2} \right)+\frac {1}{\sqrt{x^2+2}}=\frac 2 {(x^2+2)^{3/2}}$$


When I posted my answer, I just saw Yves Daoust's answer. It is much simpler that mine. However, I do not delete it since, as said, it was done for your curiosity.

  • $\begingroup$ Yep, good attempt, but unfortunately it doesn't simplify nicely. $\endgroup$ – Yves Daoust Sep 21 '16 at 7:29

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