Calculus Examples

$y = {(x + 1)}^{x + 1}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} ({(x + 1)}^{x + 1})$

Step 2

The derivative of $y$ with respect to $x$ is $y'$ .

$y'$

Step 3

Differentiate the right side of the equation.

Tap for more steps...

Step 3.1

Use the properties of logarithms to simplify the differentiation.

Tap for more steps...

Step 3.1.1

Rewrite ${(x + 1)}^{x + 1}$ as $e^{\ln ({(x + 1)}^{x + 1})}$ .

$\frac{d}{d x} [e^{\ln ({(x + 1)}^{x + 1})}]$

Step 3.1.2

Expand $\ln ({(x + 1)}^{x + 1})$ by moving $x + 1$ outside the logarithm.

$\frac{d}{d x} [e^{(x + 1) \ln (x + 1)}]$

Step 3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = (x + 1) \ln (x + 1)$ .

Tap for more steps...

Step 3.2.1

To apply the Chain Rule, set $u_{1}$ as $(x + 1) \ln (x + 1)$ .

$\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [(x + 1) \ln (x + 1)]$

Step 3.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$e^{u_{1}} \frac{d}{d x} [(x + 1) \ln (x + 1)]$

Step 3.2.3

Replace all occurrences of $u_{1}$ with $(x + 1) \ln (x + 1)$ .

$e^{(x + 1) \ln (x + 1)} \frac{d}{d x} [(x + 1) \ln (x + 1)]$

Step 3.3

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x + 1$ and $g (x) = \ln (x + 1)$ .

$e^{(x + 1) \ln (x + 1)} ((x + 1) \frac{d}{d x} [\ln (x + 1)] + \ln (x + 1) \frac{d}{d x} [x + 1])$

Step 3.4

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = x + 1$ .

Tap for more steps...

Step 3.4.1

To apply the Chain Rule, set $u_{2}$ as $x + 1$ .

$e^{(x + 1) \ln (x + 1)} ((x + 1) (\frac{d}{d u_{2}} [\ln (u_{2})] \frac{d}{d x} [x + 1]) + \ln (x + 1) \frac{d}{d x} [x + 1])$

Step 3.4.2

The derivative of $\ln (u_{2})$ with respect to $u_{2}$ is $\frac{1}{u_{2}}$ .

$e^{(x + 1) \ln (x + 1)} ((x + 1) (\frac{1}{u_{2}} \frac{d}{d x} [x + 1]) + \ln (x + 1) \frac{d}{d x} [x + 1])$

Step 3.4.3

Replace all occurrences of $u_{2}$ with $x + 1$ .

$e^{(x + 1) \ln (x + 1)} ((x + 1) (\frac{1}{x + 1} \frac{d}{d x} [x + 1]) + \ln (x + 1) \frac{d}{d x} [x + 1])$

Step 3.5

Differentiate.

Tap for more steps...

Step 3.5.1

By the Sum Rule, the derivative of $x + 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [1]$ .

$e^{(x + 1) \ln (x + 1)} ((x + 1) \frac{1}{x + 1} (\frac{d}{d x} [x] + \frac{d}{d x} [1]) + \ln (x + 1) \frac{d}{d x} [x + 1])$

Step 3.5.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$e^{(x + 1) \ln (x + 1)} ((x + 1) \frac{1}{x + 1} (1 + \frac{d}{d x} [1]) + \ln (x + 1) \frac{d}{d x} [x + 1])$

Step 3.5.3

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$e^{(x + 1) \ln (x + 1)} ((x + 1) \frac{1}{x + 1} (1 + 0) + \ln (x + 1) \frac{d}{d x} [x + 1])$

Step 3.5.4

Simplify the expression.

Tap for more steps...

Step 3.5.4.1

Add $1$ and $0$ .

$e^{(x + 1) \ln (x + 1)} ((x + 1) \frac{1}{x + 1} \cdot 1 + \ln (x + 1) \frac{d}{d x} [x + 1])$

Step 3.5.4.2

Multiply $x + 1$ by $1$ .

$e^{(x + 1) \ln (x + 1)} ((x + 1) \frac{1}{x + 1} + \ln (x + 1) \frac{d}{d x} [x + 1])$

Step 3.5.5

By the Sum Rule, the derivative of $x + 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [1]$ .

$e^{(x + 1) \ln (x + 1)} ((x + 1) \frac{1}{x + 1} + \ln (x + 1) (\frac{d}{d x} [x] + \frac{d}{d x} [1]))$

Step 3.5.6

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$e^{(x + 1) \ln (x + 1)} ((x + 1) \frac{1}{x + 1} + \ln (x + 1) (1 + \frac{d}{d x} [1]))$

Step 3.5.7

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$e^{(x + 1) \ln (x + 1)} ((x + 1) \frac{1}{x + 1} + \ln (x + 1) (1 + 0))$

Step 3.5.8

Simplify the expression.

Tap for more steps...

Step 3.5.8.1

Add $1$ and $0$ .

$e^{(x + 1) \ln (x + 1)} ((x + 1) \frac{1}{x + 1} + \ln (x + 1) \cdot 1)$

Step 3.5.8.2

Multiply $\ln (x + 1)$ by $1$ .

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

Step 4

Reform the equation by setting the left side equal to the right side.

$y' = e^{(x + 1) \ln (x + 1)} ((x + 1) (\frac{1}{x + 1}) + \ln (x + 1))$

Step 5

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = e^{(x + 1) \ln (x + 1)} ((x + 1) (\frac{1}{x + 1}) + \ln (x + 1))$