0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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}

$\endgroup$
1
$\begingroup$

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

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

$$\frac{x_h^2}{x_h^2+2}-\frac{x^2}{x^2+2}.$$

Then reducing to the common denominator, the numerator is

$$x_h^2(x^2+2)-x^2(x_h^2+2)=2x_h^2-2x^2=2h(x_h+x),$$

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

$$\frac{x_h}{\sqrt{x_h^2+2}}+\frac{x}{\sqrt{x^2+2}}\to\frac{2x}{\sqrt{x^2+2}}.$$

Putting all together,

$$f'(x)=\frac{4x}{(x^2+2)^2}\frac{\sqrt{x^2+2}}{2x}=2(x^2+2)^{-3/2}.$$

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

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

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

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

$$f'(x)=\frac{2x(1-f^2(x))}{2f(x)(x^2+2)},$$

which leads to the desired result.


Proof:

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

$\endgroup$
  • $\begingroup$ This one is really elegant ! Thanks for providing this solution. Cheers. $\endgroup$ – Claude Leibovici Sep 21 '16 at 10:38
0
$\begingroup$

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

Edit

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.

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

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.