How do you find the derivative of  f(x)= 2/(1+x^2) ?

Oct 29, 2016

$y = f \left(x\right)$

$y = \frac{2}{1 + {x}^{2}}$

${y}^{-} 1 = \frac{1 + {x}^{2}}{2}$

$2 {y}^{-} 1 = 1 + {x}^{2}$

Now use implicit differentiation...

$- 2 {y}^{-} 2 \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x$

$- \frac{2}{{y}^{2}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x$

Divide expressions on both sides of the equation by 2...

$- \frac{1}{y} ^ 2 \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = x$

Multiply expressions on both sides of the equation by -1...

$\frac{1}{y} ^ 2 \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = - x$

Now multiply expressions on both sides of the equation by y^2...

$\frac{\mathrm{dy}}{\mathrm{dx}} = - x \cdot {y}^{2}$

Don't forget the real value of y...

$\frac{\mathrm{dy}}{\mathrm{dx}} = - x \cdot {\left(\frac{2}{1 + {x}^{2}}\right)}^{2}$

Clean up the final result...

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{4 x}{{\left(1 + {x}^{2}\right)}^{2}}$

Which means that...

$f ' \left(x\right) = - \frac{4 x}{{\left(1 + {x}^{2}\right)}^{2}}$