Calculus Examples

$f (x) = e^{3 x^{3} + 1} \ln (2 x^{3} + 3)$

Step 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 $f (x) = e^{3 x^{3} + 1}$ and $g (x) = \ln (2 x^{3} + 3)$ .

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

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = 2 x^{3} + 3$ .

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

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

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

Step 3

Differentiate.

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Step 3.1

Combine $\frac{1}{2 x^{3} + 3}$ and $e^{3 x^{3} + 1}$ .

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

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

Step 3.5

Multiply $3$ by $2$ .

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

Step 3.6

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$\frac{e^{3 x^{3} + 1}}{2 x^{3} + 3} (6 x^{2} + 0) + \ln (2 x^{3} + 3) \frac{d}{d x} [e^{3 x^{3} + 1}]$

Step 3.7

Combine fractions.

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Step 3.7.1

Add $6 x^{2}$ and $0$ .

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

Step 3.7.2

Combine $6$ and $\frac{e^{3 x^{3} + 1}}{2 x^{3} + 3}$ .

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

Step 3.7.3

Combine $\frac{6 e^{3 x^{3} + 1}}{2 x^{3} + 3}$ and $x^{2}$ .

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

Step 4

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

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Step 4.1

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

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

Step 4.2

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

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

Step 4.3

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

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

Step 5

Differentiate.

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Step 5.1

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

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

Step 5.2

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

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

Step 5.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

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

Step 5.4

Multiply $3$ by $3$ .

$\frac{6 e^{3 x^{3} + 1} x^{2}}{2 x^{3} + 3} + \ln (2 x^{3} + 3) (e^{3 x^{3} + 1} (9 x^{2} + \frac{d}{d x} [1]))$

Step 5.5

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

$\frac{6 e^{3 x^{3} + 1} x^{2}}{2 x^{3} + 3} + \ln (2 x^{3} + 3) (e^{3 x^{3} + 1} (9 x^{2} + 0))$

Step 5.6

Add $9 x^{2}$ and $0$ .

$\frac{6 e^{3 x^{3} + 1} x^{2}}{2 x^{3} + 3} + \ln (2 x^{3} + 3) (e^{3 x^{3} + 1} (9 x^{2}))$

Step 6

To write $\ln (2 x^{3} + 3) (e^{3 x^{3} + 1} (9 x^{2}))$ as a fraction with a common denominator, multiply by $\frac{2 x^{3} + 3}{2 x^{3} + 3}$ .

$\frac{6 e^{3 x^{3} + 1} x^{2}}{2 x^{3} + 3} + \frac{\ln (2 x^{3} + 3) (e^{3 x^{3} + 1} (9 x^{2})) (2 x^{3} + 3)}{2 x^{3} + 3}$

Step 7

Combine the numerators over the common denominator.

$\frac{6 e^{3 x^{3} + 1} x^{2} + \ln (2 x^{3} + 3) (e^{3 x^{3} + 1} (9 x^{2})) (2 x^{3} + 3)}{2 x^{3} + 3}$

Step 8

Simplify.

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Step 8.1

Simplify the numerator.

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Step 8.1.1

Simplify each term.

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Step 8.1.1.1

Rewrite using the commutative property of multiplication.

$\frac{6 e^{3 x^{3} + 1} x^{2} + 9 \ln (2 x^{3} + 3) (e^{3 x^{3} + 1} x^{2}) (2 x^{3} + 3)}{2 x^{3} + 3}$

Step 8.1.1.2

Simplify $9 \ln (2 x^{3} + 3)$ by moving $9$ inside the logarithm.

$\frac{6 e^{3 x^{3} + 1} x^{2} + \ln ({(2 x^{3} + 3)}^{9}) (e^{3 x^{3} + 1} x^{2}) (2 x^{3} + 3)}{2 x^{3} + 3}$

Step 8.1.1.3

Apply the distributive property.

$\frac{6 e^{3 x^{3} + 1} x^{2} + \ln ({(2 x^{3} + 3)}^{9}) (e^{3 x^{3} + 1} x^{2}) (2 x^{3}) + \ln ({(2 x^{3} + 3)}^{9}) (e^{3 x^{3} + 1} x^{2}) \cdot 3}{2 x^{3} + 3}$

Step 8.1.1.4

Rewrite using the commutative property of multiplication.

$\frac{6 e^{3 x^{3} + 1} x^{2} + 2 \ln ({(2 x^{3} + 3)}^{9}) e^{3 x^{3} + 1} x^{2} x^{3} + \ln ({(2 x^{3} + 3)}^{9}) (e^{3 x^{3} + 1} x^{2}) \cdot 3}{2 x^{3} + 3}$

Step 8.1.1.5

Multiply $\ln ({(2 x^{3} + 3)}^{9}) (e^{3 x^{3} + 1} x^{2}) \cdot 3$ .

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Step 8.1.1.5.1

Reorder $\ln ({(2 x^{3} + 3)}^{9})$ and $3$ .

$\frac{6 e^{3 x^{3} + 1} x^{2} + 2 \ln ({(2 x^{3} + 3)}^{9}) e^{3 x^{3} + 1} x^{2} x^{3} + e^{3 x^{3} + 1} x^{2} \cdot (3 \cdot \ln ({(2 x^{3} + 3)}^{9}))}{2 x^{3} + 3}$

Step 8.1.1.5.2

Simplify $3 \cdot \ln ({(2 x^{3} + 3)}^{9})$ by moving $3$ inside the logarithm.

$\frac{6 e^{3 x^{3} + 1} x^{2} + 2 \ln ({(2 x^{3} + 3)}^{9}) e^{3 x^{3} + 1} x^{2} x^{3} + e^{3 x^{3} + 1} x^{2} \cdot \ln ({({(2 x^{3} + 3)}^{9})}^{3})}{2 x^{3} + 3}$

Step 8.1.1.6

Simplify each term.

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Step 8.1.1.6.1

Multiply $x^{2}$ by $x^{3}$ by adding the exponents.

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Step 8.1.1.6.1.1

Move $x^{3}$ .

$\frac{6 e^{3 x^{3} + 1} x^{2} + 2 \ln ({(2 x^{3} + 3)}^{9}) e^{3 x^{3} + 1} (x^{3} x^{2}) + e^{3 x^{3} + 1} x^{2} \cdot \ln ({({(2 x^{3} + 3)}^{9})}^{3})}{2 x^{3} + 3}$

Step 8.1.1.6.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{6 e^{3 x^{3} + 1} x^{2} + 2 \ln ({(2 x^{3} + 3)}^{9}) e^{3 x^{3} + 1} x^{3 + 2} + e^{3 x^{3} + 1} x^{2} \cdot \ln ({({(2 x^{3} + 3)}^{9})}^{3})}{2 x^{3} + 3}$

Step 8.1.1.6.1.3

Add $3$ and $2$ .

$\frac{6 e^{3 x^{3} + 1} x^{2} + 2 \ln ({(2 x^{3} + 3)}^{9}) e^{3 x^{3} + 1} x^{5} + e^{3 x^{3} + 1} x^{2} \cdot \ln ({({(2 x^{3} + 3)}^{9})}^{3})}{2 x^{3} + 3}$

Step 8.1.1.6.2

Simplify $2 \ln ({(2 x^{3} + 3)}^{9})$ by moving $2$ inside the logarithm.

$\frac{6 e^{3 x^{3} + 1} x^{2} + \ln ({({(2 x^{3} + 3)}^{9})}^{2}) e^{3 x^{3} + 1} x^{5} + e^{3 x^{3} + 1} x^{2} \cdot \ln ({({(2 x^{3} + 3)}^{9})}^{3})}{2 x^{3} + 3}$

Step 8.1.1.6.3

Multiply the exponents in ${({(2 x^{3} + 3)}^{9})}^{2}$ .

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Step 8.1.1.6.3.1

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

$\frac{6 e^{3 x^{3} + 1} x^{2} + \ln ({(2 x^{3} + 3)}^{9 \cdot 2}) e^{3 x^{3} + 1} x^{5} + e^{3 x^{3} + 1} x^{2} \cdot \ln ({({(2 x^{3} + 3)}^{9})}^{3})}{2 x^{3} + 3}$

Step 8.1.1.6.3.2

Multiply $9$ by $2$ .

$\frac{6 e^{3 x^{3} + 1} x^{2} + \ln ({(2 x^{3} + 3)}^{18}) e^{3 x^{3} + 1} x^{5} + e^{3 x^{3} + 1} x^{2} \cdot \ln ({({(2 x^{3} + 3)}^{9})}^{3})}{2 x^{3} + 3}$

Step 8.1.1.6.4

Multiply the exponents in ${({(2 x^{3} + 3)}^{9})}^{3}$ .

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Step 8.1.1.6.4.1

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

$\frac{6 e^{3 x^{3} + 1} x^{2} + \ln ({(2 x^{3} + 3)}^{18}) e^{3 x^{3} + 1} x^{5} + e^{3 x^{3} + 1} x^{2} \cdot \ln ({(2 x^{3} + 3)}^{9 \cdot 3})}{2 x^{3} + 3}$

Step 8.1.1.6.4.2

Multiply $9$ by $3$ .

$\frac{6 e^{3 x^{3} + 1} x^{2} + \ln ({(2 x^{3} + 3)}^{18}) e^{3 x^{3} + 1} x^{5} + e^{3 x^{3} + 1} x^{2} \cdot \ln ({(2 x^{3} + 3)}^{27})}{2 x^{3} + 3}$

$\frac{6 e^{3 x^{3} + 1} x^{2} + \ln ({(2 x^{3} + 3)}^{18}) e^{3 x^{3} + 1} x^{5} + e^{3 x^{3} + 1} x^{2} \ln ({(2 x^{3} + 3)}^{27})}{2 x^{3} + 3}$

Step 8.1.2

Reorder factors in $6 e^{3 x^{3} + 1} x^{2} + \ln ({(2 x^{3} + 3)}^{18}) e^{3 x^{3} + 1} x^{5} + e^{3 x^{3} + 1} x^{2} \ln ({(2 x^{3} + 3)}^{27})$ .

$\frac{6 x^{2} e^{3 x^{3} + 1} + x^{5} e^{3 x^{3} + 1} \ln ({(2 x^{3} + 3)}^{18}) + x^{2} e^{3 x^{3} + 1} \ln ({(2 x^{3} + 3)}^{27})}{2 x^{3} + 3}$

Step 8.2

Reorder terms.

$\frac{6 e^{3 x^{3} + 1} x^{2} + e^{3 x^{3} + 1} x^{5} \ln ({(2 x^{3} + 3)}^{18}) + e^{3 x^{3} + 1} x^{2} \ln ({(2 x^{3} + 3)}^{27})}{2 x^{3} + 3}$