Calculus Examples

$y = \sin (3 x + 4)$

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} (\sin (3 x + 4))$

The derivative of $y$ with respect to $x$ is $y'$ .

$y'$

Differentiate the right side of the equation.

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Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sin (x)$ and $g (x) = 3 x + 4$ .

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To apply the Chain Rule, set $u$ as $3 x + 4$ .

$\frac{d}{d u} [\sin (u)] \frac{d}{d x} [3 x + 4]$

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

$\cos (u) \frac{d}{d x} [3 x + 4]$

Replace all occurrences of $u$ with $3 x + 4$ .

$\cos (3 x + 4) \frac{d}{d x} [3 x + 4]$

Differentiate.

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

$\cos (3 x + 4) (\frac{d}{d x} [3 x] + \frac{d}{d x} [4])$

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

$\cos (3 x + 4) (3 \frac{d}{d x} [x] + \frac{d}{d x} [4])$

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

$\cos (3 x + 4) (3 \cdot 1 + \frac{d}{d x} [4])$

Multiply $3$ by $1$ .

$\cos (3 x + 4) (3 + \frac{d}{d x} [4])$

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

$\cos (3 x + 4) (3 + 0)$

Simplify the expression.

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Add $3$ and $0$ .

$\cos (3 x + 4) \cdot 3$

Move $3$ to the left of $\cos (3 x + 4)$ .

$3 \cos (3 x + 4)$

Reform the equation by setting the left side equal to the right side.

$y' = 3 \cos (3 x + 4)$

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = 3 \cos (3 x + 4)$