Calculus Examples

$\ln (3 e^{(2 x - 5)} {(3 x^{3} + 5)}^{7})$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = 3 e^{2 x - 5} {(3 x^{3} + 5)}^{7}$ .

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Step 1.1

To apply the Chain Rule, set $u_{1}$ as $3 e^{2 x - 5} {(3 x^{3} + 5)}^{7}$ .

$\frac{d}{d u_{1}} [\ln (u_{1})] \frac{d}{d x} [3 e^{2 x - 5} {(3 x^{3} + 5)}^{7}]$

Step 1.2

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

$\frac{1}{u_{1}} \frac{d}{d x} [3 e^{2 x - 5} {(3 x^{3} + 5)}^{7}]$

Step 1.3

Replace all occurrences of $u_{1}$ with $3 e^{2 x - 5} {(3 x^{3} + 5)}^{7}$ .

$\frac{1}{3 e^{2 x - 5} {(3 x^{3} + 5)}^{7}} \frac{d}{d x} [3 e^{2 x - 5} {(3 x^{3} + 5)}^{7}]$

Step 2

Differentiate using the Constant Multiple Rule.

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Step 2.1

Since $3$ is constant with respect to $x$ , the derivative of $3 e^{2 x - 5} {(3 x^{3} + 5)}^{7}$ with respect to $x$ is $3 \frac{d}{d x} [e^{2 x - 5} {(3 x^{3} + 5)}^{7}]$ .

$\frac{1}{3 e^{2 x - 5} {(3 x^{3} + 5)}^{7}} (3 \frac{d}{d x} [e^{2 x - 5} {(3 x^{3} + 5)}^{7}])$

Step 2.2

Simplify terms.

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Step 2.2.1

Combine $3$ and $\frac{1}{3 e^{2 x - 5} {(3 x^{3} + 5)}^{7}}$ .

$\frac{3}{3 e^{2 x - 5} {(3 x^{3} + 5)}^{7}} \frac{d}{d x} [e^{2 x - 5} {(3 x^{3} + 5)}^{7}]$

Step 2.2.2

Cancel the common factor of $3$ .

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Step 2.2.2.1

Cancel the common factor.

$\frac{3}{3 e^{2 x - 5} {(3 x^{3} + 5)}^{7}} \frac{d}{d x} [e^{2 x - 5} {(3 x^{3} + 5)}^{7}]$

Step 2.2.2.2

Rewrite the expression.

$\frac{1}{e^{2 x - 5} {(3 x^{3} + 5)}^{7}} \frac{d}{d x} [e^{2 x - 5} {(3 x^{3} + 5)}^{7}]$

Step 3

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 $f (x) = e^{2 x - 5}$ and $g (x) = {(3 x^{3} + 5)}^{7}$ .

$\frac{1}{e^{2 x - 5} {(3 x^{3} + 5)}^{7}} (e^{2 x - 5} \frac{d}{d x} [{(3 x^{3} + 5)}^{7}] + {(3 x^{3} + 5)}^{7} \frac{d}{d x} [e^{2 x - 5}])$

Step 4

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{7}$ and $g (x) = 3 x^{3} + 5$ .

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Step 4.1

To apply the Chain Rule, set $u_{2}$ as $3 x^{3} + 5$ .

$\frac{1}{e^{2 x - 5} {(3 x^{3} + 5)}^{7}} (e^{2 x - 5} (\frac{d}{d u_{2}} [{u_{2}}^{7}] \frac{d}{d x} [3 x^{3} + 5]) + {(3 x^{3} + 5)}^{7} \frac{d}{d x} [e^{2 x - 5}])$

Step 4.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 7$ .

$\frac{1}{e^{2 x - 5} {(3 x^{3} + 5)}^{7}} (e^{2 x - 5} (7 {u_{2}}^{6} \frac{d}{d x} [3 x^{3} + 5]) + {(3 x^{3} + 5)}^{7} \frac{d}{d x} [e^{2 x - 5}])$

Step 4.3

Replace all occurrences of $u_{2}$ with $3 x^{3} + 5$ .

$\frac{1}{e^{2 x - 5} {(3 x^{3} + 5)}^{7}} (e^{2 x - 5} (7 {(3 x^{3} + 5)}^{6} \frac{d}{d x} [3 x^{3} + 5]) + {(3 x^{3} + 5)}^{7} \frac{d}{d x} [e^{2 x - 5}])$

Step 5

Differentiate.

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Step 5.1

By the Sum Rule, the derivative of $3 x^{3} + 5$ with respect to $x$ is $\frac{d}{d x} [3 x^{3}] + \frac{d}{d x} [5]$ .

$\frac{1}{e^{2 x - 5} {(3 x^{3} + 5)}^{7}} (e^{2 x - 5} (7 {(3 x^{3} + 5)}^{6} (\frac{d}{d x} [3 x^{3}] + \frac{d}{d x} [5])) + {(3 x^{3} + 5)}^{7} \frac{d}{d x} [e^{2 x - 5}])$

Step 5.2

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

$\frac{1}{e^{2 x - 5} {(3 x^{3} + 5)}^{7}} (e^{2 x - 5} (7 {(3 x^{3} + 5)}^{6} (3 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [5])) + {(3 x^{3} + 5)}^{7} \frac{d}{d x} [e^{2 x - 5}])$

Step 5.3

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

$\frac{1}{e^{2 x - 5} {(3 x^{3} + 5)}^{7}} (e^{2 x - 5} (7 {(3 x^{3} + 5)}^{6} (3 (3 x^{2}) + \frac{d}{d x} [5])) + {(3 x^{3} + 5)}^{7} \frac{d}{d x} [e^{2 x - 5}])$

Step 5.4

Multiply $3$ by $3$ .

$\frac{1}{e^{2 x - 5} {(3 x^{3} + 5)}^{7}} (e^{2 x - 5} (7 {(3 x^{3} + 5)}^{6} (9 x^{2} + \frac{d}{d x} [5])) + {(3 x^{3} + 5)}^{7} \frac{d}{d x} [e^{2 x - 5}])$

Step 5.5

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$\frac{1}{e^{2 x - 5} {(3 x^{3} + 5)}^{7}} (e^{2 x - 5} (7 {(3 x^{3} + 5)}^{6} (9 x^{2} + 0)) + {(3 x^{3} + 5)}^{7} \frac{d}{d x} [e^{2 x - 5}])$

Step 5.6

Simplify the expression.

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Step 5.6.1

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

$\frac{1}{e^{2 x - 5} {(3 x^{3} + 5)}^{7}} (e^{2 x - 5} (7 {(3 x^{3} + 5)}^{6} (9 x^{2})) + {(3 x^{3} + 5)}^{7} \frac{d}{d x} [e^{2 x - 5}])$

Step 5.6.2

Multiply $9$ by $7$ .

$\frac{1}{e^{2 x - 5} {(3 x^{3} + 5)}^{7}} (e^{2 x - 5} (63 {(3 x^{3} + 5)}^{6} x^{2}) + {(3 x^{3} + 5)}^{7} \frac{d}{d x} [e^{2 x - 5}])$

Step 6

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 $g (x) = 2 x - 5$ .

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Step 6.1

To apply the Chain Rule, set $u_{3}$ as $2 x - 5$ .

$\frac{1}{e^{2 x - 5} {(3 x^{3} + 5)}^{7}} (e^{2 x - 5} (63 {(3 x^{3} + 5)}^{6} x^{2}) + {(3 x^{3} + 5)}^{7} (\frac{d}{d u_{3}} [e^{u_{3}}] \frac{d}{d x} [2 x - 5]))$

Step 6.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{3}} [a^{u_{3}}]$ is $a^{u_{3}} \ln (a)$ where $a$ = $e$ .

$\frac{1}{e^{2 x - 5} {(3 x^{3} + 5)}^{7}} (e^{2 x - 5} (63 {(3 x^{3} + 5)}^{6} x^{2}) + {(3 x^{3} + 5)}^{7} (e^{u_{3}} \frac{d}{d x} [2 x - 5]))$

Step 6.3

Replace all occurrences of $u_{3}$ with $2 x - 5$ .

$\frac{1}{e^{2 x - 5} {(3 x^{3} + 5)}^{7}} (e^{2 x - 5} (63 {(3 x^{3} + 5)}^{6} x^{2}) + {(3 x^{3} + 5)}^{7} (e^{2 x - 5} \frac{d}{d x} [2 x - 5]))$

Step 7

Differentiate.

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Step 7.1

By the Sum Rule, the derivative of $2 x - 5$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 5]$ .

$\frac{1}{e^{2 x - 5} {(3 x^{3} + 5)}^{7}} (e^{2 x - 5} (63 {(3 x^{3} + 5)}^{6} x^{2}) + {(3 x^{3} + 5)}^{7} e^{2 x - 5} (\frac{d}{d x} [2 x] + \frac{d}{d x} [- 5]))$

Step 7.2

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

$\frac{1}{e^{2 x - 5} {(3 x^{3} + 5)}^{7}} (e^{2 x - 5} (63 {(3 x^{3} + 5)}^{6} x^{2}) + {(3 x^{3} + 5)}^{7} e^{2 x - 5} (2 \frac{d}{d x} [x] + \frac{d}{d x} [- 5]))$

Step 7.3

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

$\frac{1}{e^{2 x - 5} {(3 x^{3} + 5)}^{7}} (e^{2 x - 5} (63 {(3 x^{3} + 5)}^{6} x^{2}) + {(3 x^{3} + 5)}^{7} e^{2 x - 5} (2 \cdot 1 + \frac{d}{d x} [- 5]))$

Step 7.4

Multiply $2$ by $1$ .

$\frac{1}{e^{2 x - 5} {(3 x^{3} + 5)}^{7}} (e^{2 x - 5} (63 {(3 x^{3} + 5)}^{6} x^{2}) + {(3 x^{3} + 5)}^{7} e^{2 x - 5} (2 + \frac{d}{d x} [- 5]))$

Step 7.5

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$\frac{1}{e^{2 x - 5} {(3 x^{3} + 5)}^{7}} (e^{2 x - 5} (63 {(3 x^{3} + 5)}^{6} x^{2}) + {(3 x^{3} + 5)}^{7} e^{2 x - 5} (2 + 0))$

Step 7.6

Simplify the expression.

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Step 7.6.1

Add $2$ and $0$ .

$\frac{1}{e^{2 x - 5} {(3 x^{3} + 5)}^{7}} (e^{2 x - 5} (63 {(3 x^{3} + 5)}^{6} x^{2}) + {(3 x^{3} + 5)}^{7} e^{2 x - 5} \cdot 2)$

Step 7.6.2

Move $2$ to the left of ${(3 x^{3} + 5)}^{7} e^{2 x - 5}$ .

$\frac{1}{e^{2 x - 5} {(3 x^{3} + 5)}^{7}} (e^{2 x - 5} (63 {(3 x^{3} + 5)}^{6} x^{2}) + 2 \cdot ({(3 x^{3} + 5)}^{7} e^{2 x - 5}))$

Step 7.6.3

Reorder the factors of $\frac{1}{e^{2 x - 5} {(3 x^{3} + 5)}^{7}} (e^{2 x - 5} (63 {(3 x^{3} + 5)}^{6} x^{2}) + 2 {(3 x^{3} + 5)}^{7} e^{2 x - 5})$ .

$(e^{2 x - 5} (63 {(3 x^{3} + 5)}^{6} x^{2}) + 2 {(3 x^{3} + 5)}^{7} e^{2 x - 5}) \frac{1}{e^{2 x - 5} {(3 x^{3} + 5)}^{7}}$