# How do you differentiate y= (3+2^x)^x?

Nov 29, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(3 + {2}^{x}\right)}^{x} \left(\frac{\ln \left(2\right) x {2}^{x}}{3 + {2}^{x}} + \ln \left(3 + {2}^{x}\right)\right)$

#### Explanation:

The simplest method is to use what is known as logarithmic differentiation. We take (Natural) logarithms of both sides and the use implicit differentiation:

$y = {\left(3 + {2}^{x}\right)}^{x}$

Taking logarithms we get:

$\ln y = \ln \left\{{\left(3 + {2}^{x}\right)}^{x}\right\}$
$\ln y = x \ln \left(3 + {2}^{x}\right)$

Then we can differentiate the LHS Implicitly, and the RHS using the product rule to get:
$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \left(x\right) \left(\frac{d}{\mathrm{dx}} \ln \left(3 + {2}^{x}\right)\right) + \left(\ln \left(3 + {2}^{x}\right)\right) \left(1\right)$
$\therefore \frac{1}{3 + {2}^{x}} ^ x \frac{\mathrm{dy}}{\mathrm{dx}} = x \frac{d}{\mathrm{dx}} \ln \left(3 + {2}^{x}\right) + \ln \left(3 + {2}^{x}\right) \ldots \left[1\right]$

To deal with $\frac{d}{\mathrm{dx}} \ln \left(3 + {2}^{x}\right)$ we use the chain rule:

$\setminus \setminus \setminus \setminus \setminus \frac{d}{\mathrm{dx}} \ln \left(3 + {2}^{x}\right) = \frac{1}{3 + {2}^{x}} \frac{d}{\mathrm{dx}} \left(3 + {2}^{x}\right)$
$\therefore \frac{d}{\mathrm{dx}} \ln \left(3 + {2}^{x}\right) = \frac{1}{3 + {2}^{x}} \ln \left(2\right) {2}^{x}$
$\therefore \frac{d}{\mathrm{dx}} \ln \left(3 + {2}^{x}\right) = \frac{\ln \left(2\right) {2}^{x}}{3 + {2}^{x}}$

Substituting the last result into[1] we get:

$\frac{1}{3 + {2}^{x}} ^ x \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\ln \left(2\right) x {2}^{x}}{3 + {2}^{x}} + \ln \left(3 + {2}^{x}\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(3 + {2}^{x}\right)}^{x} \left(\frac{\ln \left(2\right) x {2}^{x}}{3 + {2}^{x}} + \ln \left(3 + {2}^{x}\right)\right)$