Algebra Examples

${(5 x)}^{3 \cos (2 x)}$

Step 1

Simplify with factoring out.

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Step 1.1

Factor $5$ out of $5 x$ .

$\frac{d}{d x} [{(5 (x))}^{3 \cos (2 x)}]$

Step 1.2

Apply the product rule to $5 (x)$ .

$\frac{d}{d x} [5^{3 \cos (2 x)} x^{3 \cos (2 x)}]$

Step 2

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 $f (x) = 5^{3 \cos (2 x)}$ and $g (x) = x^{3 \cos (2 x)}$ .

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

Step 3

Use the properties of logarithms to simplify the differentiation.

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Step 3.1

Rewrite $x^{3 \cos (2 x)}$ as $e^{\ln (x^{3 \cos (2 x)})}$ .

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

Step 3.2

Expand $\ln (x^{3 \cos (2 x)})$ by moving $3 \cos (2 x)$ outside the logarithm.

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

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 $g (x) = 3 \cos (2 x) \ln (x)$ .

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Step 4.1

To apply the Chain Rule, set $u_{1}$ as $3 \cos (2 x) \ln (x)$ .

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

Step 4.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

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

Step 4.3

Replace all occurrences of $u_{1}$ with $3 \cos (2 x) \ln (x)$ .

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

Step 5

Differentiate using the Constant Multiple Rule.

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Step 5.1

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

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

Step 5.2

Move $3$ to the left of $5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)}$ .

$3 \cdot (5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)}) \frac{d}{d x} [\cos (2 x) \ln (x)] + x^{3 \cos (2 x)} \frac{d}{d x} [5^{3 \cos (2 x)}]$

Step 6

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

$3 \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} (\cos (2 x) \frac{d}{d x} [\ln (x)] + \ln (x) \frac{d}{d x} [\cos (2 x)]) + x^{3 \cos (2 x)} \frac{d}{d x} [5^{3 \cos (2 x)}]$

Step 7

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

$3 \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} (\cos (2 x) \frac{1}{x} + \ln (x) \frac{d}{d x} [\cos (2 x)]) + x^{3 \cos (2 x)} \frac{d}{d x} [5^{3 \cos (2 x)}]$

Step 8

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

$3 \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} (\frac{\cos (2 x)}{x} + \ln (x) \frac{d}{d x} [\cos (2 x)]) + x^{3 \cos (2 x)} \frac{d}{d x} [5^{3 \cos (2 x)}]$

Step 9

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 $g (x) = 2 x$ .

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Step 9.1

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

$3 \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} (\frac{\cos (2 x)}{x} + \ln (x) (\frac{d}{d u_{2}} [\cos (u_{2})] \frac{d}{d x} [2 x])) + x^{3 \cos (2 x)} \frac{d}{d x} [5^{3 \cos (2 x)}]$

Step 9.2

The derivative of $\cos (u_{2})$ with respect to $u_{2}$ is $- \sin (u_{2})$ .

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

Step 9.3

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

$3 \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} (\frac{\cos (2 x)}{x} + \ln (x) (- \sin (2 x) \frac{d}{d x} [2 x])) + x^{3 \cos (2 x)} \frac{d}{d x} [5^{3 \cos (2 x)}]$

Step 10

Differentiate.

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Step 10.1

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

$3 \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} (\frac{\cos (2 x)}{x} + \ln (x) (- \sin (2 x) (2 \frac{d}{d x} [x]))) + x^{3 \cos (2 x)} \frac{d}{d x} [5^{3 \cos (2 x)}]$

Step 10.2

Multiply $2$ by $- 1$ .

$3 \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} (\frac{\cos (2 x)}{x} + \ln (x) (- 2 \sin (2 x) \frac{d}{d x} [x])) + x^{3 \cos (2 x)} \frac{d}{d x} [5^{3 \cos (2 x)}]$

Step 10.3

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

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

Step 10.4

Multiply $- 2$ by $1$ .

$3 \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} (\frac{\cos (2 x)}{x} + \ln (x) (- 2 \sin (2 x))) + x^{3 \cos (2 x)} \frac{d}{d x} [5^{3 \cos (2 x)}]$

Step 11

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

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Step 11.1

To apply the Chain Rule, set $u_{3}$ as $3 \cos (2 x)$ .

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

Step 11.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{3}} [a^{u_{3}}]$ is $a^{u_{3}} \ln (a)$ where $a$ = $5$ .

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

Step 11.3

Replace all occurrences of $u_{3}$ with $3 \cos (2 x)$ .

$3 \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} (\frac{\cos (2 x)}{x} + \ln (x) (- 2 \sin (2 x))) + x^{3 \cos (2 x)} (5^{3 \cos (2 x)} \ln (5) \frac{d}{d x} [3 \cos (2 x)])$

Step 12

Differentiate using the Constant Multiple Rule.

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Step 12.1

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

$3 \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} (\frac{\cos (2 x)}{x} + \ln (x) (- 2 \sin (2 x))) + x^{3 \cos (2 x)} \cdot 5^{3 \cos (2 x)} \ln (5) (3 \frac{d}{d x} [\cos (2 x)])$

Step 12.2

Move $3$ to the left of $x^{3 \cos (2 x)} \cdot 5^{3 \cos (2 x)} \ln (5)$ .

$3 \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} (\frac{\cos (2 x)}{x} + \ln (x) (- 2 \sin (2 x))) + 3 \cdot (x^{3 \cos (2 x)} \cdot 5^{3 \cos (2 x)} \ln (5)) \frac{d}{d x} [\cos (2 x)]$

Step 13

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 $g (x) = 2 x$ .

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Step 13.1

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

$3 \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} (\frac{\cos (2 x)}{x} + \ln (x) (- 2 \sin (2 x))) + 3 x^{3 \cos (2 x)} \cdot 5^{3 \cos (2 x)} \ln (5) (\frac{d}{d u_{4}} [\cos (u_{4})] \frac{d}{d x} [2 x])$

Step 13.2

The derivative of $\cos (u_{4})$ with respect to $u_{4}$ is $- \sin (u_{4})$ .

$3 \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} (\frac{\cos (2 x)}{x} + \ln (x) (- 2 \sin (2 x))) + 3 x^{3 \cos (2 x)} \cdot 5^{3 \cos (2 x)} \ln (5) (- \sin (u_{4}) \frac{d}{d x} [2 x])$

Step 13.3

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

$3 \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} (\frac{\cos (2 x)}{x} + \ln (x) (- 2 \sin (2 x))) + 3 x^{3 \cos (2 x)} \cdot 5^{3 \cos (2 x)} \ln (5) (- \sin (2 x) \frac{d}{d x} [2 x])$

Step 14

Differentiate.

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Step 14.1

Multiply $- 1$ by $3$ .

$3 \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} (\frac{\cos (2 x)}{x} + \ln (x) (- 2 \sin (2 x))) - 3 x^{3 \cos (2 x)} \cdot 5^{3 \cos (2 x)} \ln (5) (\sin (2 x) \frac{d}{d x} [2 x])$

Step 14.2

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

$3 \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} (\frac{\cos (2 x)}{x} + \ln (x) (- 2 \sin (2 x))) - 3 x^{3 \cos (2 x)} \cdot 5^{3 \cos (2 x)} \ln (5) \sin (2 x) (2 \frac{d}{d x} [x])$

Step 14.3

Multiply $2$ by $- 3$ .

$3 \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} (\frac{\cos (2 x)}{x} + \ln (x) (- 2 \sin (2 x))) - 6 x^{3 \cos (2 x)} \cdot 5^{3 \cos (2 x)} \ln (5) \sin (2 x) \frac{d}{d x} [x]$

Step 14.4

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

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

Step 14.5

Multiply $- 6$ by $1$ .

$3 \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} (\frac{\cos (2 x)}{x} + \ln (x) (- 2 \sin (2 x))) - 6 x^{3 \cos (2 x)} \cdot 5^{3 \cos (2 x)} \ln (5) \sin (2 x)$

Step 15

Simplify.

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Step 15.1

Apply the distributive property.

$3 \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} \frac{\cos (2 x)}{x} + 3 \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} (\ln (x) (- 2 \sin (2 x))) - 6 x^{3 \cos (2 x)} \cdot 5^{3 \cos (2 x)} \ln (5) \sin (2 x)$

Step 15.2

Combine terms.

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Step 15.2.1

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

$5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} \frac{3 \cos (2 x)}{x} + 3 \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} (\ln (x) (- 2 \sin (2 x))) - 6 x^{3 \cos (2 x)} \cdot 5^{3 \cos (2 x)} \ln (5) \sin (2 x)$

Step 15.2.2

Combine $\frac{3 \cos (2 x)}{x}$ and $5^{3 \cos (2 x)}$ .

$\frac{3 \cos (2 x) \cdot 5^{3 \cos (2 x)}}{x} e^{3 \cos (2 x) \ln (x)} + 3 \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} (\ln (x) (- 2 \sin (2 x))) - 6 x^{3 \cos (2 x)} \cdot 5^{3 \cos (2 x)} \ln (5) \sin (2 x)$

Step 15.2.3

Combine $\frac{3 \cos (2 x) \cdot 5^{3 \cos (2 x)}}{x}$ and $e^{3 \cos (2 x) \ln (x)}$ .

$\frac{3 \cos (2 x) \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)}}{x} + 3 \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} (\ln (x) (- 2 \sin (2 x))) - 6 x^{3 \cos (2 x)} \cdot 5^{3 \cos (2 x)} \ln (5) \sin (2 x)$

Step 15.2.4

Multiply $- 2$ by $3$ .

$\frac{3 \cos (2 x) \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)}}{x} - 6 \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} \ln (x) \sin (2 x) - 6 x^{3 \cos (2 x)} \cdot 5^{3 \cos (2 x)} \ln (5) \sin (2 x)$

Step 15.3

Reorder terms.

$- 6 \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)} \sin (2 x) \ln (x) - 6 \cdot 5^{3 \cos (2 x)} x^{3 \cos (2 x)} \sin (2 x) \ln (5) + \frac{3 \cos (2 x) \cdot 5^{3 \cos (2 x)} e^{3 \cos (2 x) \ln (x)}}{x}$