Calculus Examples

$5 \log ({(2)}^{x + 10})$

Step 1

Since $5$ is constant with respect to $x$ , the derivative of $5 \log (2^{x + 10})$ with respect to $x$ is $5 \frac{d}{d x} [\log (2^{x + 10})]$ .

$5 \frac{d}{d x} [\log (2^{x + 10})]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \log (x)$ and $g (x) = 2^{x + 10}$ .

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as $2^{x + 10}$ .

$5 (\frac{d}{d u_{1}} [\log (u_{1})] \frac{d}{d x} [2^{x + 10}])$

Step 2.2

The derivative of $\log (u_{1})$ with respect to $u_{1}$ is $\frac{1}{u_{1} \ln (10)}$ .

$5 (\frac{1}{2^{x + 10} \ln (10)} \frac{d}{d x} [2^{x + 10}])$

Step 3

Combine $\frac{1}{2^{x + 10} \ln (10)}$ and $5$ .

$\frac{5}{2^{x + 10} \ln (10)} \frac{d}{d x} [2^{x + 10}]$

Step 4

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = 2^{x}$ and $g (x) = x + 10$ .

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Step 4.1

To apply the Chain Rule, set $u_{2}$ as $x + 10$ .

$\frac{5}{2^{x + 10} \ln (10)} (\frac{d}{d u_{2}} [2^{u_{2}}] \frac{d}{d x} [x + 10])$

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 $a$ = $2$ .

$\frac{5}{2^{x + 10} \ln (10)} (2^{u_{2}} \ln (2) \frac{d}{d x} [x + 10])$

Step 4.3

Replace all occurrences of $u_{2}$ with $x + 10$ .

$\frac{5}{2^{x + 10} \ln (10)} (2^{x + 10} \ln (2) \frac{d}{d x} [x + 10])$

Step 5

Differentiate.

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Step 5.1

Combine $2^{x + 10}$ and $\frac{5}{2^{x + 10} \ln (10)}$ .

$\frac{2^{x + 10} \cdot 5}{2^{x + 10} \ln (10)} (\ln (2) \frac{d}{d x} [x + 10])$

Step 5.2

Simplify terms.

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Step 5.2.1

Combine $\ln (2)$ and $\frac{2^{x + 10} \cdot 5}{2^{x + 10} \ln (10)}$ .

$\frac{\ln (2) (2^{x + 10} \cdot 5)}{2^{x + 10} \ln (10)} \frac{d}{d x} [x + 10]$

Step 5.2.2

Reorder.

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Step 5.2.2.1

Move $5$ to the left of $2^{x + 10}$ .

$\frac{\ln (2) (5 \cdot 2^{x + 10})}{2^{x + 10} \ln (10)} \frac{d}{d x} [x + 10]$

Step 5.2.2.2

Move $5$ to the left of $\ln (2)$ .

$\frac{5 \cdot \ln (2) \cdot 2^{x + 10}}{2^{x + 10} \ln (10)} \frac{d}{d x} [x + 10]$

Step 5.2.3

Cancel the common factor of $2^{x + 10}$ .

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Step 5.2.3.1

Cancel the common factor.

$\frac{5 \ln (2) \cdot 2^{x + 10}}{2^{x + 10} \ln (10)} \frac{d}{d x} [x + 10]$

Step 5.2.3.2

Rewrite the expression.

$\frac{5 \ln (2)}{\ln (10)} \frac{d}{d x} [x + 10]$

Step 5.3

By the Sum Rule, the derivative of $x + 10$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [10]$ .

$\frac{5 \ln (2)}{\ln (10)} (\frac{d}{d x} [x] + \frac{d}{d x} [10])$

Step 5.4

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

$\frac{5 \ln (2)}{\ln (10)} (1 + \frac{d}{d x} [10])$

Step 5.5

Since $10$ is constant with respect to $x$ , the derivative of $10$ with respect to $x$ is $0$ .

$\frac{5 \ln (2)}{\ln (10)} (1 + 0)$

Step 5.6

Simplify the expression.

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Step 5.6.1

Add $1$ and $0$ .

$\frac{5 \ln (2)}{\ln (10)} \cdot 1$

Step 5.6.2

Multiply $\frac{5 \ln (2)}{\ln (10)}$ by $1$ .

$\frac{5 \ln (2)}{\ln (10)}$

Step 6

Simplify the numerator.

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Step 6.1

Simplify $5 \ln (2)$ by moving $5$ inside the logarithm.

$\frac{\ln (2^{5})}{\ln (10)}$

Step 6.2

Raise $2$ to the power of $5$ .

$\frac{\ln (32)}{\ln (10)}$