Calculus Examples

$y = \arctan (- 4 x)$ , $x = 3$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \arctan (x)$ and $g (x) = - 4 x$ .

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Step 1.1

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

$\frac{d}{d u} [\arctan (u)] \frac{d}{d x} [- 4 x]$

Step 1.2

The derivative of $\arctan (u)$ with respect to $u$ is $\frac{1}{1 + u^{2}}$ .

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

Step 1.3

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

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

Step 2

Differentiate.

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Step 2.1

Factor $- 4$ out of $- 4 x$ .

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

Step 2.2

Simplify the expression.

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Step 2.2.1

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

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

Step 2.2.2

Raise $- 4$ to the power of $2$ .

$\frac{1}{1 + 16 x^{2}} \frac{d}{d x} [- 4 x]$

Step 2.3

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

$\frac{1}{1 + 16 x^{2}} (- 4 \frac{d}{d x} [x])$

Step 2.4

Combine fractions.

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Step 2.4.1

Combine $- 4$ and $\frac{1}{1 + 16 x^{2}}$ .

$\frac{- 4}{1 + 16 x^{2}} \frac{d}{d x} [x]$

Step 2.4.2

Move the negative in front of the fraction.

$- \frac{4}{1 + 16 x^{2}} \frac{d}{d x} [x]$

Step 2.5

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

$- \frac{4}{1 + 16 x^{2}} \cdot 1$

Step 2.6

Multiply $- 1$ by $1$ .

$- \frac{4}{1 + 16 x^{2}}$

Step 3

Evaluate the derivative at $x = 3$ .

$- \frac{4}{1 + 16 {(3)}^{2}}$

Step 4

Simplify the denominator.

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Step 4.1

Raise $3$ to the power of $2$ .

$- \frac{4}{1 + 16 \cdot 9}$

Step 4.2

Multiply $16$ by $9$ .

$- \frac{4}{1 + 144}$

Step 4.3

Add $1$ and $144$ .

$- \frac{4}{145}$