Elasticity of the function

Question

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 ? $$E_xf(x_0) = \frac{f'(x_0)·x_0} {f(x_0)} $$

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

$$E_xf(x) = \frac{\frac{2ax} {(1+x^2)^2}} {\frac{ax^2} {1+ x^2}} = \frac{2} {1+x^2} $$

$$E_xf(10) = \frac{2} {1+10^2} = \frac{2} {101} $$

Answer 1

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.