Calculus Examples

$w = \log_{8} (2^{p} - 1)$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d p} (w) = \frac{d}{d p} (\log_{8} (2^{p} - 1))$

Step 2

The derivative of $w$ with respect to $p$ is $w'$ .

$w'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Differentiate using the chain rule, which states that $\frac{d}{d p} [f (g (p))]$ is $f' (g (p)) g' (p)$ where $f (p) = \log_{8} (p)$ and $g (p) = 2^{p} - 1$ .

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Step 3.1.1

To apply the Chain Rule, set $u$ as $2^{p} - 1$ .

$\frac{d}{d u} [\log_{8} (u)] \frac{d}{d p} [2^{p} - 1]$

Step 3.1.2

The derivative of $\log_{8} (u)$ with respect to $u$ is $\frac{1}{u \ln (8)}$ .

$\frac{1}{u \ln (8)} \frac{d}{d p} [2^{p} - 1]$

Step 3.1.3

Replace all occurrences of $u$ with $2^{p} - 1$ .

$\frac{1}{(2^{p} - 1) \ln (8)} \frac{d}{d p} [2^{p} - 1]$

Step 3.2

By the Sum Rule, the derivative of $2^{p} - 1$ with respect to $p$ is $\frac{d}{d p} [2^{p}] + \frac{d}{d p} [- 1]$ .

$\frac{1}{(2^{p} - 1) \ln (8)} (\frac{d}{d p} [2^{p}] + \frac{d}{d p} [- 1])$

Step 3.3

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

$\frac{1}{(2^{p} - 1) \ln (8)} (2^{p} \ln (2) + \frac{d}{d p} [- 1])$

Step 3.4

Differentiate using the Constant Rule.

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Step 3.4.1

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

$\frac{1}{(2^{p} - 1) \ln (8)} (2^{p} \ln (2) + 0)$

Step 3.4.2

Combine fractions.

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Step 3.4.2.1

Add $2^{p} \ln (2)$ and $0$ .

$\frac{1}{(2^{p} - 1) \ln (8)} (2^{p} \ln (2))$

Step 3.4.2.2

Combine $2^{p}$ and $\frac{1}{(2^{p} - 1) \ln (8)}$ .

$\frac{2^{p}}{(2^{p} - 1) \ln (8)} \ln (2)$

Step 3.4.2.3

Combine $\frac{2^{p}}{(2^{p} - 1) \ln (8)}$ and $\ln (2)$ .

$\frac{2^{p} \ln (2)}{(2^{p} - 1) \ln (8)}$

Step 3.5

Simplify.

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Step 3.5.1

Apply the distributive property.

$\frac{2^{p} \ln (2)}{2^{p} \ln (8) - 1 \ln (8)}$

Step 3.5.2

Rewrite $- 1 \ln (8)$ as $- \ln (8)$ .

$\frac{2^{p} \ln (2)}{2^{p} \ln (8) - \ln (8)}$

Step 4

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

$w' = \frac{2^{p} \ln (2)}{2^{p} \ln (8) - \ln (8)}$

Step 5

Replace $w'$ with $\frac{d w}{d p}$ .

$\frac{d w}{d p} = \frac{2^{p} \ln (2)}{2^{p} \ln (8) - \ln (8)}$