Calculus Examples

$\frac{f (2 + h) - f (2)}{h}$

Step 1

Differentiate using the Quotient Rule which states that $\frac{d}{d h} [\frac{f (h)}{g (h)}]$ is $\frac{g (h) \frac{d}{d h} [f (h)] - f (h) \frac{d}{d h} [g (h)]}{{g (h)}^{2}}$ where $f (h) = f (2 + h) - f (2)$ and $g (h) = h$ .

$\frac{h \frac{d}{d h} [f (2 + h) - f (2)] - (f (2 + h) - f (2)) \frac{d}{d h} [h]}{h^{2}}$

Step 2

Differentiate.

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Step 2.1

By the Sum Rule, the derivative of $f (2 + h) - f (2)$ with respect to $h$ is $\frac{d}{d h} [f (2 + h)] + \frac{d}{d h} [- f (2)]$ .

$\frac{h (\frac{d}{d h} [f (2 + h)] + \frac{d}{d h} [- f (2)]) - (f (2 + h) - f (2)) \frac{d}{d h} [h]}{h^{2}}$

Step 2.2

Since $- f (2)$ is constant with respect to $h$ , the derivative of $- f (2)$ with respect to $h$ is $0$ .

$\frac{h (\frac{d}{d h} [f (2 + h)] + 0) - (f (2 + h) - f (2)) \frac{d}{d h} [h]}{h^{2}}$

Step 2.3

Add $\frac{d}{d h} [f (2 + h)]$ and $0$ .

$\frac{h \frac{d}{d h} [f (2 + h)] - (f (2 + h) - f (2)) \frac{d}{d h} [h]}{h^{2}}$

Step 2.4

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

$\frac{h \frac{d}{d h} [f (2 + h)] - (f (2 + h) - f (2)) \cdot 1}{h^{2}}$

Step 2.5

Multiply $- 1$ by $1$ .

$\frac{h \frac{d}{d h} [f (2 + h)] - (f (2 + h) - f (2))}{h^{2}}$

Step 3

Simplify.

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Step 3.1

Apply the distributive property.

$\frac{h \frac{d}{d h} [f (2 + h)] - f (2 + h) - - f (2)}{h^{2}}$

Step 3.2

Multiply $- - f (2)$ .

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Step 3.2.1

Multiply $- 1$ by $- 1$ .

$\frac{h \frac{d}{d h} [f (2 + h)] - f (2 + h) + 1 f (2)}{h^{2}}$

Step 3.2.2

Multiply $f (2)$ by $1$ .

$\frac{h \frac{d}{d h} [f (2 + h)] - f (2 + h) + f (2)}{h^{2}}$

Step 3.3

Reorder terms.

$\frac{h \frac{d}{d h} [f (2 + h)] + f (2) - f (2 + h)}{h^{2}}$