Calculus Examples

$f (x) = 2 x + 2$

Consider the limit definition of the derivative.

$f' (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

Find the components of the definition.

Tap for more steps...

Evaluate the function at $x = x + h$ .

Tap for more steps...

Replace the variable $x$ with $x + h$ in the expression.

$f (x + h) = 2 (x + h) + 2$

Simplify the result.

Tap for more steps...

Apply the distributive property.

$f (x + h) = 2 x + 2 h + 2$

The final answer is $2 x + 2 h + 2$ .

$2 x + 2 h + 2$

Find the components of the definition.

$f (x + h) = 2 x + 2 h + 2$

$f (x) = 2 x + 2$

$f (x + h) = 2 x + 2 h + 2$

$f (x) = 2 x + 2$

Plug in the components.

$f' (x) = lim_{h \to 0} \frac{2 x + 2 h + 2 - (2 x + 2)}{h}$

Simplify.

Tap for more steps...

Simplify the numerator.

Tap for more steps...

Group the last three terms.

$f' (x) = lim_{h \to 0} \frac{2 x + (2 h + (2 - (2 x + 2)))}{h}$

Factor.

Tap for more steps...

Factor $2$ out of $2 x + 2$ .

Tap for more steps...

Factor $2$ out of $2 x$ .

$f' (x) = lim_{h \to 0} \frac{2 x + (2 h + (2 - (2 (x) + 2)))}{h}$

Factor $2$ out of $2$ .

$f' (x) = lim_{h \to 0} \frac{2 x + (2 h + (2 - (2 (x) + 2 (1))))}{h}$

Factor $2$ out of $2 (x) + 2 (1)$ .

$f' (x) = lim_{h \to 0} \frac{2 x + (2 h + (2 - (2 (x + 1))))}{h}$

Remove unnecessary parentheses.

$f' (x) = lim_{h \to 0} \frac{2 x + (2 h + (2 - 1 \cdot (2 (x + 1))))}{h}$

Multiply $- 1$ by $2$ .

$f' (x) = lim_{h \to 0} \frac{2 x + (2 h + (2 - 2 (x + 1)))}{h}$

Remove unnecessary parentheses.

$f' (x) = lim_{h \to 0} \frac{2 x + 2 h + 2 - 2 (x + 1)}{h}$

Factor $2$ out of $2 x + 2 h + 2 - 2 (x + 1)$ .

Tap for more steps...

Factor $2$ out of $2 x$ .

$f' (x) = lim_{h \to 0} \frac{2 (x) + 2 h + 2 - 2 (x + 1)}{h}$

Factor $2$ out of $2 h$ .

$f' (x) = lim_{h \to 0} \frac{2 (x) + 2 h + 2 - 2 (x + 1)}{h}$

Factor $2$ out of $2$ .

$f' (x) = lim_{h \to 0} \frac{2 (x) + 2 h + 2 (1) - 2 (x + 1)}{h}$

Factor $2$ out of $- 2 (x + 1)$ .

$f' (x) = lim_{h \to 0} \frac{2 (x) + 2 h + 2 (1) + 2 (- (x + 1))}{h}$

Factor $2$ out of $2 (x) + 2 h$ .

$f' (x) = lim_{h \to 0} \frac{2 (x + h) + 2 (1) + 2 (- (x + 1))}{h}$

Factor $2$ out of $2 (x + h) + 2 (1)$ .

$f' (x) = lim_{h \to 0} \frac{2 (x + h + 1) + 2 (- (x + 1))}{h}$

Factor $2$ out of $2 (x + h + 1) + 2 (- (x + 1))$ .

$f' (x) = lim_{h \to 0} \frac{2 (x + h + 1 - (x + 1))}{h}$

Apply the distributive property.

$f' (x) = lim_{h \to 0} \frac{2 (x + h + 1 + (- x - 1 \cdot 1))}{h}$

Multiply $- 1$ by $1$ .

$f' (x) = lim_{h \to 0} \frac{2 (x + h + 1 + (- x - 1))}{h}$

Remove unnecessary parentheses.

$f' (x) = lim_{h \to 0} \frac{2 (x + h + 1 - x - 1)}{h}$

Subtract $x$ from $x$ .

$f' (x) = lim_{h \to 0} \frac{2 (h + 1 + 0 - 1)}{h}$

Add $h$ and $0$ .

$f' (x) = lim_{h \to 0} \frac{2 (h + 1 - 1)}{h}$

Subtract $1$ from $1$ .

$f' (x) = lim_{h \to 0} \frac{2 (h + 0)}{h}$

Add $h$ and $0$ .

$f' (x) = lim_{h \to 0} \frac{2 h}{h}$

Reduce the expression by cancelling the common factors.

Tap for more steps...

Cancel the common factor.

$f' (x) = lim_{h \to 0} \frac{2 h}{h}$

Divide $2$ by $1$ .

$f' (x) = lim_{h \to 0} 2$

Evaluate the limit of $2$ which is constant as $h$ approaches $0$ .

$2$

Enter YOUR Problem