# How do you differentiate y=ln(e^x+sqrt(1+e^(2x)))?

Aug 9, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{x}}{\sqrt{1 + {e}^{2 x}}}$

#### Explanation:

Use the chain rule.

$u \left(x\right) = {e}^{x} + {\left(1 + {e}^{2 x}\right)}^{\frac{1}{2}} \mathmr{and} y = \ln \left(u\right)$

$\frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{u} = \frac{1}{{e}^{x} + {\left(1 + {e}^{2 x}\right)}^{\frac{1}{2}}}$

$\frac{\mathrm{du}}{\mathrm{dx}} = {e}^{x} + \frac{d}{\mathrm{dx}} \left({\left(1 + {e}^{2 x}\right)}^{\frac{1}{2}}\right)$

For the square root use chain rule again with

$\phi = {\left(1 + {e}^{2 x}\right)}^{\frac{1}{2}}$

$v \left(x\right) = 1 + {e}^{2 x} \mathmr{and} \phi = {v}^{\frac{1}{2}}$

$\frac{\mathrm{dv}}{\mathrm{dx}} = 2 {e}^{2 x} \mathmr{and} \frac{\mathrm{dp} h i}{\mathrm{dv}} = \frac{1}{2 \sqrt{v}}$

$\frac{\mathrm{dp} h i}{\mathrm{dx}} = \frac{\mathrm{dp} h i}{\mathrm{dv}} \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{{e}^{2 x}}{\sqrt{1 + {e}^{2 x}}}$

$\therefore \frac{\mathrm{du}}{\mathrm{dx}} = {e}^{x} + \frac{{e}^{2 x}}{\sqrt{1 + {e}^{2 x}}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

$= \frac{1}{{e}^{x} + {\left(1 + {e}^{2 x}\right)}^{\frac{1}{2}}} \cdot \left({e}^{x} + \frac{{e}^{2 x}}{\sqrt{1 + {e}^{2 x}}}\right)$

=e^x/(e^x + sqrt(1+e^(2x))) + e^(2x)/(sqrt(1 + e^(2x))(e^x+sqrt(1+e^(2x)))

Bringing together over LCD:

=(e^xsqrt(1+e^(2x)) + e^(2x))/(sqrt(1 + e^(2x))(e^x+sqrt(1+e^(2x)))

Take factor of ${e}^{x}$ out of numerator:

=(e^x(sqrt(1+e^(2x)) + e^x))/(sqrt(1 + e^(2x))(e^x+sqrt(1+e^(2x)))

Cancel out and obtain

$= \frac{{e}^{x}}{\sqrt{1 + {e}^{2 x}}}$