# How do you differentiate ((e^x)+x)^(1/x)?

Feb 16, 2017

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

#### Explanation:

The best approach here is to use logarithmic differentiation to remove the variable exponent, as follows:

Let $y = {\left({e}^{x} + x\right)}^{\frac{1}{x}}$

$\implies \ln \setminus y = \ln \setminus \left\{{\left({e}^{x} + x\right)}^{\frac{1}{x}}\right\}$
$\text{ } = \frac{1}{x} \setminus \ln \left({e}^{x} + x\right)$

We can now differentiate the LHS implicitly, and the RHS using the product rule to get:

$\setminus \setminus \setminus \setminus \setminus \frac{1}{y} \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{1}{x}\right) \left(\frac{d}{\mathrm{dx}} \ln \left({e}^{x} + x\right)\right) + \left(\frac{d}{\mathrm{dx}} \frac{1}{x}\right) \left(\ln \left({e}^{x} + x\right)\right)$
 :. 1/y \ dy/dx = (1/x)( 1/(e^x+x) * (e^x+1))) +(-1/x^2)(ln(e^x+x))
$\therefore \frac{1}{y} \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{x} + 1}{x \left({e}^{x} + x\right)} - \frac{\ln \left({e}^{x} + x\right)}{x} ^ 2$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = {\left({e}^{x} + x\right)}^{\frac{1}{x}} \left\{\frac{{e}^{x} + 1}{x \left({e}^{x} + x\right)} - \frac{\ln \left({e}^{x} + x\right)}{x} ^ 2\right\}$