Calculus Examples

$6 e^{4 x} - 2 e^{3 x} - 4$

Step 1

By the Sum Rule, the derivative of $6 e^{4 x} - 2 e^{3 x} - 4$ with respect to $x$ is $\frac{d}{d x} [6 e^{4 x}] + \frac{d}{d x} [- 2 e^{3 x}] + \frac{d}{d x} [- 4]$ .

$\frac{d}{d x} [6 e^{4 x}] + \frac{d}{d x} [- 2 e^{3 x}] + \frac{d}{d x} [- 4]$

Step 2

Evaluate $\frac{d}{d x} [6 e^{4 x}]$ .

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Step 2.1

Since $6$ is constant with respect to $x$ , the derivative of $6 e^{4 x}$ with respect to $x$ is $6 \frac{d}{d x} [e^{4 x}]$ .

$6 \frac{d}{d x} [e^{4 x}] + \frac{d}{d x} [- 2 e^{3 x}] + \frac{d}{d x} [- 4]$

Step 2.2

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 $g (x) = 4 x$ .

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Step 2.2.1

To apply the Chain Rule, set $u_{1}$ as $4 x$ .

$6 (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [4 x]) + \frac{d}{d x} [- 2 e^{3 x}] + \frac{d}{d x} [- 4]$

Step 2.2.2

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

$6 (e^{u_{1}} \frac{d}{d x} [4 x]) + \frac{d}{d x} [- 2 e^{3 x}] + \frac{d}{d x} [- 4]$

Step 2.2.3

Replace all occurrences of $u_{1}$ with $4 x$ .

$6 (e^{4 x} \frac{d}{d x} [4 x]) + \frac{d}{d x} [- 2 e^{3 x}] + \frac{d}{d x} [- 4]$

Step 2.3

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

$6 (e^{4 x} (4 \frac{d}{d x} [x])) + \frac{d}{d x} [- 2 e^{3 x}] + \frac{d}{d x} [- 4]$

Step 2.4

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

$6 (e^{4 x} (4 \cdot 1)) + \frac{d}{d x} [- 2 e^{3 x}] + \frac{d}{d x} [- 4]$

Step 2.5

Multiply $4$ by $1$ .

$6 (e^{4 x} \cdot 4) + \frac{d}{d x} [- 2 e^{3 x}] + \frac{d}{d x} [- 4]$

Step 2.6

Move $4$ to the left of $e^{4 x}$ .

$6 (4 \cdot e^{4 x}) + \frac{d}{d x} [- 2 e^{3 x}] + \frac{d}{d x} [- 4]$

Step 2.7

Multiply $4$ by $6$ .

$24 e^{4 x} + \frac{d}{d x} [- 2 e^{3 x}] + \frac{d}{d x} [- 4]$

Step 3

Evaluate $\frac{d}{d x} [- 2 e^{3 x}]$ .

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Step 3.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 e^{3 x}$ with respect to $x$ is $- 2 \frac{d}{d x} [e^{3 x}]$ .

$24 e^{4 x} - 2 \frac{d}{d x} [e^{3 x}] + \frac{d}{d x} [- 4]$

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 $g (x) = 3 x$ .

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Step 3.2.1

To apply the Chain Rule, set $u_{2}$ as $3 x$ .

$24 e^{4 x} - 2 (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [3 x]) + \frac{d}{d x} [- 4]$

Step 3.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$24 e^{4 x} - 2 (e^{u_{2}} \frac{d}{d x} [3 x]) + \frac{d}{d x} [- 4]$

Step 3.2.3

Replace all occurrences of $u_{2}$ with $3 x$ .

$24 e^{4 x} - 2 (e^{3 x} \frac{d}{d x} [3 x]) + \frac{d}{d x} [- 4]$

Step 3.3

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$24 e^{4 x} - 2 (e^{3 x} (3 \frac{d}{d x} [x])) + \frac{d}{d x} [- 4]$

Step 3.4

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

$24 e^{4 x} - 2 (e^{3 x} (3 \cdot 1)) + \frac{d}{d x} [- 4]$

Step 3.5

Multiply $3$ by $1$ .

$24 e^{4 x} - 2 (e^{3 x} \cdot 3) + \frac{d}{d x} [- 4]$

Step 3.6

Move $3$ to the left of $e^{3 x}$ .

$24 e^{4 x} - 2 (3 \cdot e^{3 x}) + \frac{d}{d x} [- 4]$

Step 3.7

Multiply $3$ by $- 2$ .

$24 e^{4 x} - 6 e^{3 x} + \frac{d}{d x} [- 4]$

Step 4

Differentiate using the Constant Rule.

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Step 4.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$24 e^{4 x} - 6 e^{3 x} + 0$

Step 4.2

Add $24 e^{4 x} - 6 e^{3 x}$ and $0$ .

$24 e^{4 x} - 6 e^{3 x}$