Calculus Examples

$f (x) = 3 \sqrt{4 x} + e^{- 2 x}$

Step 1

By the Sum Rule, the derivative of $3 \sqrt{4 x} + e^{- 2 x}$ with respect to $x$ is $\frac{d}{d x} [3 \sqrt{4 x}] + \frac{d}{d x} [e^{- 2 x}]$ .

$\frac{d}{d x} [3 \sqrt{4 x}] + \frac{d}{d x} [e^{- 2 x}]$

Step 2

Evaluate $\frac{d}{d x} [3 \sqrt{4 x}]$ .

Tap for more steps...

Step 2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{4 x}$ as ${(4 x)}^{\frac{1}{2}}$ .

$\frac{d}{d x} [3 {(4 x)}^{\frac{1}{2}}] + \frac{d}{d x} [e^{- 2 x}]$

Step 2.2

Factor $4$ out of $4 x$ .

$\frac{d}{d x} [3 {(4 (x))}^{\frac{1}{2}}] + \frac{d}{d x} [e^{- 2 x}]$

Step 2.3

Apply the product rule to $4 (x)$ .

$\frac{d}{d x} [3 (4^{\frac{1}{2}} x^{\frac{1}{2}})] + \frac{d}{d x} [e^{- 2 x}]$

Step 2.4

Rewrite $4$ as $2^{2}$ .

$\frac{d}{d x} [3 ({(2^{2})}^{\frac{1}{2}} x^{\frac{1}{2}})] + \frac{d}{d x} [e^{- 2 x}]$

Step 2.5

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{d}{d x} [3 (2^{2 (\frac{1}{2})} x^{\frac{1}{2}})] + \frac{d}{d x} [e^{- 2 x}]$

Step 2.6

Cancel the common factor of $2$ .

Tap for more steps...

Step 2.6.1

Cancel the common factor.

$\frac{d}{d x} [3 (2^{2 (\frac{1}{2})} x^{\frac{1}{2}})] + \frac{d}{d x} [e^{- 2 x}]$

Step 2.6.2

Rewrite the expression.

$\frac{d}{d x} [3 (2^{1} x^{\frac{1}{2}})] + \frac{d}{d x} [e^{- 2 x}]$

Step 2.7

Evaluate the exponent.

$\frac{d}{d x} [3 (2 x^{\frac{1}{2}})] + \frac{d}{d x} [e^{- 2 x}]$

Step 2.8

Multiply $2$ by $3$ .

$\frac{d}{d x} [6 x^{\frac{1}{2}}] + \frac{d}{d x} [e^{- 2 x}]$

Step 2.9

Since $6$ is constant with respect to $x$ , the derivative of $6 x^{\frac{1}{2}}$ with respect to $x$ is $6 \frac{d}{d x} [x^{\frac{1}{2}}]$ .

$6 \frac{d}{d x} [x^{\frac{1}{2}}] + \frac{d}{d x} [e^{- 2 x}]$

Step 2.10

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

$6 (\frac{1}{2} x^{\frac{1}{2} - 1}) + \frac{d}{d x} [e^{- 2 x}]$

Step 2.11

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

$6 (\frac{1}{2} x^{\frac{1}{2} - 1 \cdot \frac{2}{2}}) + \frac{d}{d x} [e^{- 2 x}]$

Step 2.12

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

$6 (\frac{1}{2} x^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}}) + \frac{d}{d x} [e^{- 2 x}]$

Step 2.13

Combine the numerators over the common denominator.

$6 (\frac{1}{2} x^{\frac{1 - 1 \cdot 2}{2}}) + \frac{d}{d x} [e^{- 2 x}]$

Step 2.14

Simplify the numerator.

Tap for more steps...

Step 2.14.1

Multiply $- 1$ by $2$ .

$6 (\frac{1}{2} x^{\frac{1 - 2}{2}}) + \frac{d}{d x} [e^{- 2 x}]$

Step 2.14.2

Subtract $2$ from $1$ .

$6 (\frac{1}{2} x^{\frac{- 1}{2}}) + \frac{d}{d x} [e^{- 2 x}]$

Step 2.15

Move the negative in front of the fraction.

$6 (\frac{1}{2} x^{- \frac{1}{2}}) + \frac{d}{d x} [e^{- 2 x}]$

Step 2.16

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

$6 \frac{x^{- \frac{1}{2}}}{2} + \frac{d}{d x} [e^{- 2 x}]$

Step 2.17

Combine $6$ and $\frac{x^{- \frac{1}{2}}}{2}$ .

$\frac{6 x^{- \frac{1}{2}}}{2} + \frac{d}{d x} [e^{- 2 x}]$

Step 2.18

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

$\frac{6}{2 x^{\frac{1}{2}}} + \frac{d}{d x} [e^{- 2 x}]$

Step 2.19

Factor $2$ out of $6$ .

$\frac{2 \cdot 3}{2 x^{\frac{1}{2}}} + \frac{d}{d x} [e^{- 2 x}]$

Step 2.20

Cancel the common factors.

Tap for more steps...

Step 2.20.1

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

$\frac{2 \cdot 3}{2 (x^{\frac{1}{2}})} + \frac{d}{d x} [e^{- 2 x}]$

Step 2.20.2

Cancel the common factor.

$\frac{2 \cdot 3}{2 x^{\frac{1}{2}}} + \frac{d}{d x} [e^{- 2 x}]$

Step 2.20.3

Rewrite the expression.

$\frac{3}{x^{\frac{1}{2}}} + \frac{d}{d x} [e^{- 2 x}]$

Step 3

Evaluate $\frac{d}{d x} [e^{- 2 x}]$ .

Tap for more steps...

Step 3.1

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 $g (x) = - 2 x$ .

Tap for more steps...

Step 3.1.1

To apply the Chain Rule, set $u$ as $- 2 x$ .

$\frac{3}{x^{\frac{1}{2}}} + \frac{d}{d u} [e^{u}] \frac{d}{d x} [- 2 x]$

Step 3.1.2

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

$\frac{3}{x^{\frac{1}{2}}} + e^{u} \frac{d}{d x} [- 2 x]$

Step 3.1.3

Replace all occurrences of $u$ with $- 2 x$ .

$\frac{3}{x^{\frac{1}{2}}} + e^{- 2 x} \frac{d}{d x} [- 2 x]$

Step 3.2

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

$\frac{3}{x^{\frac{1}{2}}} + e^{- 2 x} (- 2 \frac{d}{d x} [x])$

Step 3.3

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

$\frac{3}{x^{\frac{1}{2}}} + e^{- 2 x} (- 2 \cdot 1)$

Step 3.4

Multiply $- 2$ by $1$ .

$\frac{3}{x^{\frac{1}{2}}} + e^{- 2 x} \cdot - 2$

Step 3.5

Move $- 2$ to the left of $e^{- 2 x}$ .

$\frac{3}{x^{\frac{1}{2}}} - 2 e^{- 2 x}$