# How to solve this derivative? with steps

##  \frac(1)(e^x + \sqrt {e^x + 1}

Jun 4, 2018

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

#### Explanation:

Writing your term in the form

${\left({e}^{x} + \sqrt{{e}^{x} + 1}\right)}^{- 1}$ so we get after the rule

$\left({x}^{n}\right) ' = n {x}^{n - 1}$
$- {\left({e}^{x} + \sqrt{{e}^{x} + 1}\right)}^{- 2} \cdot \left({e}^{x} + {e}^{x} / \left(2 \sqrt{{e}^{x} + 1}\right)\right)$
the second factor Comes from the chain rule, if we
differentiate $\left({e}^{x} + \sqrt{{e}^{x} + 1}\right)$ with respect to $x$

Jun 4, 2018

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

#### Explanation:

Take it in small steps. break the problem down into pieces, and solve those, then put everything together

Start with the formula for finding the derivative of the quotient of 2 functions:

$\frac{d}{\mathrm{dx}} \frac{f \left(x\right)}{g} \left(x\right) = \frac{f ' \left(x\right) \cdot g \left(x\right) - f \left(x\right) \cdot g ' \left(x\right)}{g} {\left(x\right)}^{2}$

...this problem allows at least some simplification since the numerator is the constant 1, so $f ' \left(x\right) = 0$

So, if we denote the denominator in the original function as g(x), the derivative is:

$\frac{0 - g ' \left(x\right)}{g {\left(x\right)}^{2}}$ (call this equation 1)

We have to then find $g ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({e}^{x} + \sqrt{{e}^{x} + 1}\right)$

We know that $\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$, but we have to find $\frac{d}{\mathrm{dx}} \sqrt{{e}^{x} + 1}$

...rewrite the square root term as ${\left({e}^{x} + 1\right)}^{\frac{1}{2}}$
...you find this using the chain rule, and the rule for finding derivatives of powers of functions:

$\frac{d}{\mathrm{dx}} \left(f {\left(x\right)}^{n}\right) = n f {\left(x\right)}^{n - 1} \cdot f ' \left(x\right)$

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

so, going back to $g \left(x\right) = {e}^{x} + \sqrt{{e}^{x} + 1}$, then

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

...you can rewrite this as:

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

and now, all we have to do is find ${\left({e}^{x} + \sqrt{{e}^{x} + 1}\right)}^{2}$:

...and that's just a simple binomial multiplication:

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

So now, you can put all this together:

$\frac{0 - g ' \left(x\right)}{g {\left(x\right)}^{2}} = - \left(\frac{{e}^{x} + {e}^{x} / \left(2 \cdot \sqrt{{e}^{x} + 1}\right)}{{e}^{2 x} + 2 \left({e}^{x} \cdot \sqrt{{e}^{x} + 1}\right) + \left({e}^{x} + 1\right)}\right)$

GOOD LUCK