Calculus Examples

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

Step 1

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

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

Step 2

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

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

Multiply $2$ by $3$ .

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

Step 3

Evaluate $\frac{d}{d x} [2 \cos (x)]$ .

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Step 3.1

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

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

Step 3.2

The derivative of $\cos (x)$ with respect to $x$ is $- \sin (x)$ .

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

Step 3.3

Multiply $- 1$ by $2$ .

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

Step 4

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

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Step 4.1

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

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

Step 4.2

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 4.2.1

To apply the Chain Rule, set $u$ as $4 x$ .

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

Step 4.2.2

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

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

Step 4.2.3

Replace all occurrences of $u$ with $4 x$ .

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

Step 4.3

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

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

Step 4.4

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

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

Step 4.5

Multiply $4$ by $1$ .

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

Step 4.6

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

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

Step 4.7

Multiply $4$ by $- 4$ .

$6 x - 2 \sin (x) - 16 e^{4 x} + \frac{d}{d x} [π]$

Step 5

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

$6 x - 2 \sin (x) - 16 e^{4 x} + 0$

Step 6

Simplify.

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Step 6.1

Add $6 x - 2 \sin (x) - 16 e^{4 x}$ and $0$ .

$6 x - 2 \sin (x) - 16 e^{4 x}$

Step 6.2

Reorder terms.

$6 x - 16 e^{4 x} - 2 \sin (x)$