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

Question

Solving the equality

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

could help to defined the proof for the question 1.3 of the book Neural Networks by Simon S. Haykin (a simplified text).

Let

$$\sigma(v) = \frac{v}{\sqrt(1 + v^2)}$$

show that the derivative of $\sigma(v)$ is given by

$$\frac{d\sigma}{dv} = \frac{\sigma^3(v)}{v^3}$$

I wrote an alternative resolution for that question in this post, but another insights and approaches could help to improve the final anwser.

Answer 1

$$\dfrac{1}{\sqrt{1+x^2}}-\dfrac{x^2}{{(1+x^2)}^{\frac{3}{2}}}$$ Let,$$p=\sqrt{1+x^2}$$ So,we get $$ \begin{array} \\L.H.S&=&\\ &=& \dfrac{1}{p}-\dfrac{x^2}{p^3}\\ &=& \dfrac{p^2-x^2}{p^3}\\ &=& \dfrac{1+x^2-x^2}{{(1+x^2)}^{\frac{3}{2}}}\\ &=& \dfrac{1}{{(1+x^2)}^{\frac{3}{2}}}=R.H.S_{[proved]}\\ \end{array} $$

Answer 2

Hint: $$\dfrac{1}{\sqrt{1 + x^2}} = \dfrac{1}{(1 + x^2)^{1/2}} = \dfrac{1+x^2}{(1+x^2)(1 + x^2)^{1/2}} = \dfrac{1+x^2}{(1 + x^2)^{3/2}}$$

Answer 3

It suffices simplify algebrically for example by $1+x^2=y$ we have

$$\frac1{\sqrt y}-\frac{y-1}{y\sqrt{y}}=\frac{\sqrt y}y-\frac{(y-1)\sqrt y}{y^2}=\frac{y\sqrt y-y\sqrt y+\sqrt y}{y^2}=\ldots$$

or as a tricky alternative just use $x=\sinh y \implies 1+x^2=\cosh^2 y$ to obtain

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

$$\dfrac{1}{\cosh y} - \dfrac{\sinh^2y}{\cosh^3 y} = \dfrac{1}{\cosh^3 y}$$

$$\dfrac{\cosh^2 y-\sinh^2 y}{\cosh^3 y} = \dfrac{1}{\cosh^3 y}$$ $$\dfrac{1}{\cosh^3 y} = \dfrac{1}{\cosh^3 y}$$

Answer 4

I solved that problem using factorization multiplying the left part by $\frac{(1 + x^2)}{(1 + x^2)}$.

$$\dfrac{1}{\sqrt(1 + x^2)} - \dfrac{x^2}{(1 + x^2)^{3/2}}$$

$$\dfrac{1}{\sqrt(1 + x^2)} \times \dfrac{1 + x^2}{1 + x^2} - \dfrac{x^2}{(1 + x^2)^{3/2}}$$

$$\dfrac{1 + x^2}{(1 + x^2)^{1/2} \times (1 + x^2)^{1}} - \dfrac{x^2}{(1 + x^2)^{3/2}}$$

$$\dfrac{1 + x^2}{(1 + x^2)^{3/2}} - \dfrac{x^2}{(1 + x^2)^{3/2}}$$

$$\dfrac{1 + x^2 - x^2}{(1 + x^2)^{3/2}}$$

$$\dfrac{1}{(1 + x^2)^{3/2}}$$