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I am asked to find the second derivative of the function:

$$h(x)=\sqrt{x^2+1}$$ $$h(x)=(x^2+1)^\frac{1}{2}$$ $$h'=\frac{1}{2}(x^2+1)^\frac{-1}{2} 2x$$ $$h'=\frac{x}{\sqrt{x^2+1}}$$

$$h''=\frac{\sqrt{x^2+1} - x(\frac{1}{2}(x^2+1)^\frac{-1}{2}2x}{(\sqrt{x^2+1})^2}$$

And this is as far as I get. My question is regarding how to proceed algebraically to solve this question. I am having hard time seeing what the next step is in simplifying this equation.

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simplify the numerator and factorize $\frac 1{\sqrt{x^2+1}}$ in the numerator. – Raymond Manzoni Jul 22 '12 at 7:52
up vote 2 down vote accepted

After the most obvious simplifications you have

$$\begin{align*} h''(x)&=\frac{\sqrt{x^2+1} - x\left(\frac{1}{2}(x^2+1)^{-\frac12}(2x)\right)}{(\sqrt{x^2+1})^2}\\ &=\frac{\sqrt{x^2+1}-x^2(x^2+1)^{-\frac12}}{x^2+1}\\ &=\frac{\sqrt{x^2+1}-\dfrac{x^2}{\sqrt{x^2+1}}}{x^2+1}\;. \end{align*}$$

You can now either multiply the fraction directly by $\frac{\sqrt{x^2+1}}{\sqrt{x^2+1}}$ or continue simplifying the numerator. If you do the latter, you get

$$\begin{align*} \frac{\sqrt{x^2+1}-\dfrac{x^2}{\sqrt{x^2+1}}}{x^2+1}&=\frac{\dfrac{x^2+1-x^2}{\sqrt{x^2+1}}}{x^2+1}\\ &=\frac{\dfrac1{\sqrt{x^2+1}}}{x^2+1}\\ &=\frac1{(x^2+1)\sqrt{x^2+1}}\\ &=\frac1{(x^2+1)^{3/2}}\\ &=(x^2+1)^{-\frac32}\;; \end{align*}$$

if you multiply by $\frac{\sqrt{x^2+1}}{\sqrt{x^2+1}}$ directly, you end up with the same result after similar calculations.

Note that you could also have used the product rule to calculate $h''(x)$, leaving $h'(x)$ in the form $x(x^2+1)^{-1/2}$. Then

$$\begin{align*} h''(x)&=x\left(-\frac12(x^2+1)^{-3/2}(2x)\right)+(x^2+1)^{-1/2}\\ &=-\frac{x^2}{(x^2+1)^{3/2}}+\frac1{(x^2+1)^{1/2}}\\ &=-\frac{x^2}{(x^2+1)^{3/2}}+\frac{x^2+1}{(x^2+1)^{3/2}}\\ &=\frac1{(x^2+1)^{3/2}}\;, \end{align*}$$

with perhaps a little less work.

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A slightly non-algebraic way would be to use the right triangle $(1,x,\sqrt{1+x^{2}})$. If you call the angle between sides of length 1 and hypotenuse $\theta$, the derivative reduces to $$ \frac{\frac{d}{d \theta}\sin{\theta}}{\frac{d}{d \theta}\tan{\theta}} $$ This reduces to $$ \frac{\cos{\theta}}{1+\tan^{2}{\theta}} = \frac{1}{{(1+x^{2})}^{3/2}} $$

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I like this method of finding the solution. Can you elaborate on how you did this, if possible. I don't quite see the connection between the original statement and the application of trig principles. – Kurt Jul 22 '12 at 23:38
Basically when you see terms like $\sqrt{1+x^{2}}$, you can think of them as sides of a right-triangle. If you could express the problem as a trig. derivative, more often than not it becomes easier - as these are well known. I went ahead from the expression you got for the first derivative (which can be done the same way, as well). – kashyap1123 Jul 23 '12 at 0:51

There's a much, much easier and more elegant way to do this. $$ \begin{align*} h^2 = x^2+1 &\implies hh' = x\qquad(1)\\ &\implies h'^2 + h h'' = 1\\ &\implies (h h')^2 + h^3h'' = h^2 = x^2+1 \\ &\implies h^3 h'' = 1\\ \end{align*} $$ The last relation follows from $(1)$. QED

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This looks like an interesting way to solve this question. However, I do not understand the connections you are making. – Kurt Jul 22 '12 at 20:34
If you don't understand something, just ask! Which lines don't you understand? The first two $\implies$ are just using the chain rule to implicitly differentiate $h^2$. The third $\implies$ is just multiplying through by $h^2$. Then we substitute $(1)$ to obtain $x^2 + h^3 h''=h^2$. Now we use the definition of $h$ to cancel out the $x^2$ from each side and obtain the final result. – pre-kidney Jul 24 '12 at 21:49

$$h'' = \frac{\sqrt{x^2+1} - x^2 \frac{1}{\sqrt{x^2+1}}}{x^2+1} = \frac{\sqrt{x^2+1}}{\sqrt{x^2+1}} \frac{\sqrt{x^2+1} - x^2 \frac{1}{\sqrt{x^2+1}}}{x^2+1} = \frac{x^2+1-x^2}{\sqrt{x^2+1} ( x^2+1)}$$ $$h'' = \frac{1}{\sqrt{x^2+1} ( x^2+1)} = \frac{\sqrt{x^2+1}}{x^4+2x^2+1}$$

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