Calculus Examples

$e^{2 x^{2} - x + \frac{1}{x}}$

Step 1

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 $g (x) = 2 x^{2} - x + \frac{1}{x}$ .

Tap for more steps...

Step 1.1

To apply the Chain Rule, set $u$ as $2 x^{2} - x + \frac{1}{x}$ .

$\frac{d}{d u} [e^{u}] \frac{d}{d x} [2 x^{2} - x + \frac{1}{x}]$

Step 1.2

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

$e^{u} \frac{d}{d x} [2 x^{2} - x + \frac{1}{x}]$

Step 1.3

Replace all occurrences of $u$ with $2 x^{2} - x + \frac{1}{x}$ .

$e^{2 x^{2} - x + \frac{1}{x}} \frac{d}{d x} [2 x^{2} - x + \frac{1}{x}]$

Step 2

Differentiate.

Tap for more steps...

Step 2.1

By the Sum Rule, the derivative of $2 x^{2} - x + \frac{1}{x}$ with respect to $x$ is $\frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [- x] + \frac{d}{d x} [\frac{1}{x}]$ .

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

Step 2.2

Since $2$ is constant with respect to $x$ , the derivative of $2 x^{2}$ with respect to $x$ is $2 \frac{d}{d x} [x^{2}]$ .

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

Step 2.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

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

Step 2.4

Multiply $2$ by $2$ .

$e^{2 x^{2} - x + \frac{1}{x}} (4 x + \frac{d}{d x} [- x] + \frac{d}{d x} [\frac{1}{x}])$

Step 2.5

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$e^{2 x^{2} - x + \frac{1}{x}} (4 x - \frac{d}{d x} [x] + \frac{d}{d x} [\frac{1}{x}])$

Step 2.6

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

$e^{2 x^{2} - x + \frac{1}{x}} (4 x - 1 \cdot 1 + \frac{d}{d x} [\frac{1}{x}])$

Step 2.7

Simplify the expression.

Tap for more steps...

Step 2.7.1

Multiply $- 1$ by $1$ .

$e^{2 x^{2} - x + \frac{1}{x}} (4 x - 1 + \frac{d}{d x} [\frac{1}{x}])$

Step 2.7.2

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

$e^{2 x^{2} - x + \frac{1}{x}} (4 x - 1 + \frac{d}{d x} [x^{- 1}])$

Step 2.8

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

$e^{2 x^{2} - x + \frac{1}{x}} (4 x - 1 - x^{- 2})$

Step 3

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$e^{2 x^{2} - x + \frac{1}{x}} (4 x - 1 - \frac{1}{x^{2}})$