Calculus Examples

$y = \ln (3 x + 1) \cos (2 x + 4)$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} (\ln (3 x + 1) \cos (2 x + 4))$

Step 2

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

$y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = \ln (3 x + 1)$ and $g (x) = \cos (2 x + 4)$ .

$\ln (3 x + 1) \frac{d}{d x} [\cos (2 x + 4)] + \cos (2 x + 4) \frac{d}{d x} [\ln (3 x + 1)]$

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 $f (x) = \cos (x)$ and $g (x) = 2 x + 4$ .

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Step 3.2.1

To apply the Chain Rule, set $u_{1}$ as $2 x + 4$ .

$\ln (3 x + 1) (\frac{d}{d u_{1}} [\cos (u_{1})] \frac{d}{d x} [2 x + 4]) + \cos (2 x + 4) \frac{d}{d x} [\ln (3 x + 1)]$

Step 3.2.2

The derivative of $\cos (u_{1})$ with respect to $u_{1}$ is $- \sin (u_{1})$ .

$\ln (3 x + 1) (- \sin (u_{1}) \frac{d}{d x} [2 x + 4]) + \cos (2 x + 4) \frac{d}{d x} [\ln (3 x + 1)]$

Step 3.2.3

Replace all occurrences of $u_{1}$ with $2 x + 4$ .

$\ln (3 x + 1) (- \sin (2 x + 4) \frac{d}{d x} [2 x + 4]) + \cos (2 x + 4) \frac{d}{d x} [\ln (3 x + 1)]$

Step 3.3

Differentiate.

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Step 3.3.1

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

$\ln (3 x + 1) (- \sin (2 x + 4) (\frac{d}{d x} [2 x] + \frac{d}{d x} [4])) + \cos (2 x + 4) \frac{d}{d x} [\ln (3 x + 1)]$

Step 3.3.2

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

$\ln (3 x + 1) (- \sin (2 x + 4) (2 \frac{d}{d x} [x] + \frac{d}{d x} [4])) + \cos (2 x + 4) \frac{d}{d x} [\ln (3 x + 1)]$

Step 3.3.3

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

$\ln (3 x + 1) (- \sin (2 x + 4) (2 \cdot 1 + \frac{d}{d x} [4])) + \cos (2 x + 4) \frac{d}{d x} [\ln (3 x + 1)]$

Step 3.3.4

Multiply $2$ by $1$ .

$\ln (3 x + 1) (- \sin (2 x + 4) (2 + \frac{d}{d x} [4])) + \cos (2 x + 4) \frac{d}{d x} [\ln (3 x + 1)]$

Step 3.3.5

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

$\ln (3 x + 1) (- \sin (2 x + 4) (2 + 0)) + \cos (2 x + 4) \frac{d}{d x} [\ln (3 x + 1)]$

Step 3.3.6

Simplify the expression.

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Step 3.3.6.1

Add $2$ and $0$ .

$\ln (3 x + 1) (- \sin (2 x + 4) \cdot 2) + \cos (2 x + 4) \frac{d}{d x} [\ln (3 x + 1)]$

Step 3.3.6.2

Multiply $2$ by $- 1$ .

$\ln (3 x + 1) (- 2 \sin (2 x + 4)) + \cos (2 x + 4) \frac{d}{d x} [\ln (3 x + 1)]$

Step 3.4

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = 3 x + 1$ .

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Step 3.4.1

To apply the Chain Rule, set $u_{2}$ as $3 x + 1$ .

$\ln (3 x + 1) (- 2 \sin (2 x + 4)) + \cos (2 x + 4) (\frac{d}{d u_{2}} [\ln (u_{2})] \frac{d}{d x} [3 x + 1])$

Step 3.4.2

The derivative of $\ln (u_{2})$ with respect to $u_{2}$ is $\frac{1}{u_{2}}$ .

$\ln (3 x + 1) (- 2 \sin (2 x + 4)) + \cos (2 x + 4) (\frac{1}{u_{2}} \frac{d}{d x} [3 x + 1])$

Step 3.4.3

Replace all occurrences of $u_{2}$ with $3 x + 1$ .

$\ln (3 x + 1) (- 2 \sin (2 x + 4)) + \cos (2 x + 4) (\frac{1}{3 x + 1} \frac{d}{d x} [3 x + 1])$

Step 3.5

Differentiate.

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Step 3.5.1

Combine $\frac{1}{3 x + 1}$ and $\cos (2 x + 4)$ .

$\ln (3 x + 1) (- 2 \sin (2 x + 4)) + \frac{\cos (2 x + 4)}{3 x + 1} \frac{d}{d x} [3 x + 1]$

Step 3.5.2

By the Sum Rule, the derivative of $3 x + 1$ with respect to $x$ is $\frac{d}{d x} [3 x] + \frac{d}{d x} [1]$ .

$\ln (3 x + 1) (- 2 \sin (2 x + 4)) + \frac{\cos (2 x + 4)}{3 x + 1} (\frac{d}{d x} [3 x] + \frac{d}{d x} [1])$

Step 3.5.3

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

$\ln (3 x + 1) (- 2 \sin (2 x + 4)) + \frac{\cos (2 x + 4)}{3 x + 1} (3 \frac{d}{d x} [x] + \frac{d}{d x} [1])$

Step 3.5.4

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

$\ln (3 x + 1) (- 2 \sin (2 x + 4)) + \frac{\cos (2 x + 4)}{3 x + 1} (3 \cdot 1 + \frac{d}{d x} [1])$

Step 3.5.5

Multiply $3$ by $1$ .

$\ln (3 x + 1) (- 2 \sin (2 x + 4)) + \frac{\cos (2 x + 4)}{3 x + 1} (3 + \frac{d}{d x} [1])$

Step 3.5.6

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$\ln (3 x + 1) (- 2 \sin (2 x + 4)) + \frac{\cos (2 x + 4)}{3 x + 1} (3 + 0)$

Step 3.5.7

Combine fractions.

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Step 3.5.7.1

Add $3$ and $0$ .

$\ln (3 x + 1) (- 2 \sin (2 x + 4)) + \frac{\cos (2 x + 4)}{3 x + 1} \cdot 3$

Step 3.5.7.2

Combine $\frac{\cos (2 x + 4)}{3 x + 1}$ and $3$ .

$\ln (3 x + 1) (- 2 \sin (2 x + 4)) + \frac{\cos (2 x + 4) \cdot 3}{3 x + 1}$

Step 3.5.7.3

Move $3$ to the left of $\cos (2 x + 4)$ .

$\ln (3 x + 1) (- 2 \sin (2 x + 4)) + \frac{3 \cdot \cos (2 x + 4)}{3 x + 1}$

Step 3.6

To write $\ln (3 x + 1) (- 2 \sin (2 x + 4))$ as a fraction with a common denominator, multiply by $\frac{3 x + 1}{3 x + 1}$ .

$\frac{\ln (3 x + 1) (- 2 \sin (2 x + 4)) (3 x + 1)}{3 x + 1} + \frac{3 \cos (2 x + 4)}{3 x + 1}$

Step 3.7

Combine the numerators over the common denominator.

$\frac{\ln (3 x + 1) (- 2 \sin (2 x + 4)) (3 x + 1) + 3 \cos (2 x + 4)}{3 x + 1}$

Step 3.8

Simplify.

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Step 3.8.1

Simplify the numerator.

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Step 3.8.1.1

Simplify each term.

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Step 3.8.1.1.1

Rewrite using the commutative property of multiplication.

$\frac{- 2 \ln (3 x + 1) \sin (2 x + 4) (3 x + 1) + 3 \cos (2 x + 4)}{3 x + 1}$

Step 3.8.1.1.2

Simplify $- 2 \ln (3 x + 1)$ by moving $2$ inside the logarithm.

$\frac{- \ln ({(3 x + 1)}^{2}) \sin (2 x + 4) (3 x + 1) + 3 \cos (2 x + 4)}{3 x + 1}$

Step 3.8.1.1.3

Apply the distributive property.

$\frac{- \ln ({(3 x + 1)}^{2}) \sin (2 x + 4) (3 x) - \ln ({(3 x + 1)}^{2}) \sin (2 x + 4) \cdot 1 + 3 \cos (2 x + 4)}{3 x + 1}$

Step 3.8.1.1.4

Multiply $- \ln ({(3 x + 1)}^{2}) \sin (2 x + 4) (3 x)$ .

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Step 3.8.1.1.4.1

Multiply $3$ by $- 1$ .

$\frac{- 3 \ln ({(3 x + 1)}^{2}) \sin (2 x + 4) x - \ln ({(3 x + 1)}^{2}) \sin (2 x + 4) \cdot 1 + 3 \cos (2 x + 4)}{3 x + 1}$

Step 3.8.1.1.4.2

Simplify $- 3 \ln ({(3 x + 1)}^{2})$ by moving $3$ inside the logarithm.

$\frac{- \ln ({({(3 x + 1)}^{2})}^{3}) \sin (2 x + 4) x - \ln ({(3 x + 1)}^{2}) \sin (2 x + 4) \cdot 1 + 3 \cos (2 x + 4)}{3 x + 1}$

Step 3.8.1.1.5

Multiply $- 1$ by $1$ .

$\frac{- \ln ({({(3 x + 1)}^{2})}^{3}) \sin (2 x + 4) x - \ln ({(3 x + 1)}^{2}) \sin (2 x + 4) + 3 \cos (2 x + 4)}{3 x + 1}$

Step 3.8.1.1.6

Multiply the exponents in ${({(3 x + 1)}^{2})}^{3}$ .

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Step 3.8.1.1.6.1

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

$\frac{- \ln ({(3 x + 1)}^{2 \cdot 3}) \sin (2 x + 4) x - \ln ({(3 x + 1)}^{2}) \sin (2 x + 4) + 3 \cos (2 x + 4)}{3 x + 1}$

Step 3.8.1.1.6.2

Multiply $2$ by $3$ .

$\frac{- \ln ({(3 x + 1)}^{6}) \sin (2 x + 4) x - \ln ({(3 x + 1)}^{2}) \sin (2 x + 4) + 3 \cos (2 x + 4)}{3 x + 1}$

Step 3.8.1.2

Reorder factors in $- \ln ({(3 x + 1)}^{6}) \sin (2 x + 4) x - \ln ({(3 x + 1)}^{2}) \sin (2 x + 4) + 3 \cos (2 x + 4)$ .

$\frac{- x \ln ({(3 x + 1)}^{6}) \sin (2 x + 4) - \ln ({(3 x + 1)}^{2}) \sin (2 x + 4) + 3 \cos (2 x + 4)}{3 x + 1}$

Step 3.8.2

Reorder terms.

$\frac{- x \sin (2 x + 4) \ln ({(3 x + 1)}^{6}) - \sin (2 x + 4) \ln ({(3 x + 1)}^{2}) + 3 \cos (2 x + 4)}{3 x + 1}$

Step 4

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

$y' = \frac{- x \sin (2 x + 4) \ln ({(3 x + 1)}^{6}) - \sin (2 x + 4) \ln ({(3 x + 1)}^{2}) + 3 \cos (2 x + 4)}{3 x + 1}$

Step 5

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

$\frac{d y}{d x} = \frac{- x \sin (2 x + 4) \ln ({(3 x + 1)}^{6}) - \sin (2 x + 4) \ln ({(3 x + 1)}^{2}) + 3 \cos (2 x + 4)}{3 x + 1}$