Elasticity of the function Please find the elasticity of the function f(x) in x0:
$$f(x) = \frac{ax^2} {1+ x^2}  $$ $$ x0=10$$
Could you please check is done correct ?
$$E_xf(x_0) = \frac{f'(x_0)·x_0} {f(x_0)} $$
$$f'(x)=\frac{2ax·(1+x^2)-2ax^2·2x} {(1+x^2)^2} = \frac{2ax} {(1+x^2)^2} $$
$$E_xf(x) = \frac{\frac{2ax} {(1+x^2)^2}} {\frac{ax^2} {1+ x^2}} = \frac{2} {1+x^2} $$
$$E_xf(10) = \frac{2} {1+10^2} = \frac{2} {101} $$

As you have mentioned before, the function for elasticity of a function with respect to $x$ is
$$E_x(f(x)) = \frac{f'(x) \cdot x}{f(x)}.$$
We see by the quotient rule that
$$f'(x) = \frac{(2ax)(1+x^2) - (ax^2)(2x)}{(1+x^2)^2}.$$
Thus,
$$E_x(f(x)) = \frac{\frac{(2ax)(1+x^2) - (ax^2)(2x)}{(1+x^2)^2} \cdot x}{\frac{ax^2}{1+x^2}} = \frac{2ax^2(1+x^2) - (ax^3)(2x)}{(ax^2)(1+x^2)}$$
$$=\frac{2(1+x^2) - (x)(2x)}{1+x^2} = \frac{2}{1+x^2}$$
$$\implies E_x(f(10)) = \frac{2}{1+(10)^2} = \frac{2}{101}.$$
Hence, you are correct.
