# What is the derivative of (x+1)^x?

Jan 31, 2016

#### Answer:

${\left(x + 1\right)}^{x} \left(\frac{x}{x + 1} + \ln x\right)$

#### Explanation:

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

lets use $\ln$ in the two sides of the equation,

$\ln y = x \ln \left(x + 1\right)$

now lets differentiate the two sides in respect of $x$

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = x \frac{d}{\mathrm{dx}} \left(\ln \left(x + 1\right)\right) + \ln x \frac{d}{\mathrm{dx}} \left(x\right)$

$\mathmr{and} , \frac{1}{y} \frac{d}{\mathrm{dx}} = x \cdot \frac{1}{x + 1} \frac{d}{\mathrm{dx}} \left(x + 1\right) + \ln x$

$\mathmr{and} , \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{x + 1} + \ln x$

$\mathmr{and} , \frac{\mathrm{dy}}{\mathrm{dx}} = y \left(\frac{x}{x + 1} + \ln x\right)$

$\mathmr{and} , \frac{\mathrm{dy}}{\mathrm{dx}} = {\left(x + 1\right)}^{x} \left(\frac{x}{x + 1} + \ln x\right)$

Jan 31, 2016

#### Answer:

${\left(x + 1\right)}^{x} \left(\ln \left(x + 1\right) + \frac{x}{x + 1}\right)$

#### Explanation:

I think the easiest way to do these kinds of problems (with a variable power) without memorizing the formula is through implicit differentiation.

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

Take the natural logarithm of both sides.

$\ln \left(y\right) = \ln \left({\left(x + 1\right)}^{x}\right)$

Rewrite using logarithm rules.

$\ln \left(y\right) = x \ln \left(x + 1\right)$

Differentiate both sides with respect to $x$.

The left side will spit out a $\frac{\mathrm{dy}}{\mathrm{dx}}$ term, thanks to the chain rule. The right side will require the product rule.

$\frac{1}{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \ln \left(x + 1\right) \frac{d}{\mathrm{dx}} \left(x\right) + x \frac{d}{\mathrm{dx}} \left(\ln \left(x + 1\right)\right)$

Note that differentiating $\ln \left(x + 1\right)$ requires chain rule as well, but since $\frac{d}{\mathrm{dx}} \left(x + 1\right) = 1$ it doesn't make a visible difference.

$\frac{1}{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \ln \left(x + 1\right) + x \left(\frac{1}{x + 1}\right) \frac{d}{\mathrm{dx}} \left(x + 1\right)$

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

Now, solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$ by multiplying both sides by $y$. However, remember that $y = {\left(x + 1\right)}^{x}$.

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