# How do you differentiate f(x)=4sqrt((x^2+1)/(x^2-1))?

Jun 17, 2016

$f ' \left(x\right) = \frac{- 8 x}{{\left({x}^{2} - 1\right)}^{\frac{3}{2}} {\left({x}^{2} + 1\right)}^{\frac{1}{2}}} .$

#### Explanation:

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

Taking ln of both sides,
$\ln y = \ln 4 + \frac{1}{2} \left\{\ln \left({x}^{2} + 1\right) - \ln \left({x}^{2} - 1\right)\right\}$

Diff. both sides w.r.t x,
$\frac{d}{\mathrm{dx}} \ln y = \frac{d}{\mathrm{dx}} \ln 4 + \frac{1}{2} \left\{\frac{d}{\mathrm{dx}} \ln \left({x}^{2} + 1\right) - \frac{d}{\mathrm{dx}} \ln \left({x}^{2} - 1\right)\right\} .$
$\frac{d}{\mathrm{dy}} \ln y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 0 + \frac{1}{2} \left\{\frac{1}{{x}^{2} + 1} \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right) - \frac{1}{{x}^{2} - 1} \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} - 1\right)\right\}$
1/y*dy/dx=1/2[1/(x^2+1)*2x-1/(x^2-1)*2x}=1/2*2x[{x^2-1-x^2-1]/{(x^2-1)(x^2+1)]=-(2x)/{(x^2-1)(x^2+1)}

Therefore, $\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right) = y \cdot \frac{- 2 x}{\left({x}^{2} - 1\right) \left({x}^{2} + 1\right)} = 4 \sqrt{\frac{{x}^{2} + 1}{{x}^{2} - 1}} \cdot \frac{- 2 x}{\left({x}^{2} - 1\right) \left({x}^{2} + 1\right)} = \frac{- 8 x}{{\left({x}^{2} - 1\right)}^{\frac{3}{2}} {\left({x}^{2} + 1\right)}^{\frac{1}{2}}} .$