Calculus Examples

$y = e^{4 \sqrt{x} + x^{2}}$

Step 1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

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

Step 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 $g (x) = 4 x^{\frac{1}{2}} + x^{2}$ .

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Step 2.1

To apply the Chain Rule, set $u$ as $4 x^{\frac{1}{2}} + x^{2}$ .

$\frac{d}{d u} [e^{u}] \frac{d}{d x} [4 x^{\frac{1}{2}} + x^{2}]$

Step 2.2

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

$e^{u} \frac{d}{d x} [4 x^{\frac{1}{2}} + x^{2}]$

Step 2.3

Replace all occurrences of $u$ with $4 x^{\frac{1}{2}} + x^{2}$ .

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

Step 3

Differentiate.

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Step 3.1

By the Sum Rule, the derivative of $4 x^{\frac{1}{2}} + x^{2}$ with respect to $x$ is $\frac{d}{d x} [4 x^{\frac{1}{2}}] + \frac{d}{d x} [x^{2}]$ .

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

Step 3.2

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{\frac{1}{2}}$ with respect to $x$ is $4 \frac{d}{d x} [x^{\frac{1}{2}}]$ .

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

Step 3.3

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

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

Step 4

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 5

Combine $- 1$ and $\frac{2}{2}$ .

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

Step 6

Combine the numerators over the common denominator.

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

Step 7

Simplify the numerator.

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Step 7.1

Multiply $- 1$ by $2$ .

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

Step 7.2

Subtract $2$ from $1$ .

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

Step 8

Move the negative in front of the fraction.

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

Step 9

Combine $\frac{1}{2}$ and $x^{- \frac{1}{2}}$ .

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

Step 10

Combine $4$ and $\frac{x^{- \frac{1}{2}}}{2}$ .

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

Step 11

Move $x^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 12

Factor $2$ out of $4$ .

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

Step 13

Cancel the common factors.

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Step 13.1

Factor $2$ out of $2 x^{\frac{1}{2}}$ .

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

Step 13.2

Cancel the common factor.

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

Step 13.3

Rewrite the expression.

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

Step 14

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

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