# How do you use the limit definition to find the derivative of f(x)=2/(x+4)?

Oct 21, 2016

$f ' \left(x\right) = \frac{- 2}{{\left(x + 4\right)}^{2}}$

#### Explanation:

def of derivative

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

Substitution

$f ' \left(x\right) = \lim h \to 0 \frac{\frac{2}{x + h + 4} - \frac{2}{x + 4}}{h}$

Common Denominator

$f ' \left(x\right) = \lim h \to 0 \frac{\frac{2 \left(x + 4\right)}{\left(x + 4\right) \left(x + h + 4\right)} - \frac{2 \left(x + h + 4\right)}{\left(x + 4\right) \left(x + h + 4\right)}}{h}$

Distribute and write as a single numerator

$f ' \left(x\right) = \lim h \to 0 \frac{\frac{2 x + 8}{\left(x + 4\right) \left(x + h + 4\right)} - \frac{2 x + 2 h + 8}{\left(x + 4\right) \left(x + h + 4\right)}}{h}$

$f ' \left(x\right) = \lim h \to 0 \frac{\frac{2 x + 8 - 2 x - 2 h - 8}{\left(x + 4\right) \left(x + h + 4\right)}}{h}$

Simplify

$f ' \left(x\right) = \lim h \to 0 \frac{\frac{\cancel{2 x} \cancel{+ 8} \cancel{- 2 x} - 2 h \cancel{- 8}}{\left(x + 4\right) \left(x + h + 4\right)}}{h}$

$f ' \left(x\right) = \lim h \to 0 \frac{\frac{- 2 h}{\left(x + 4\right) \left(x + h + 4\right)}}{h}$

Multiply by the reciprocal

$f ' \left(x\right) = \lim h \to 0 \frac{- 2 h}{\left(x + 4\right) \left(x + h + 4\right)} \cdot \left(\frac{1}{h}\right)$

$f ' \left(x\right) = \lim h \to 0 \frac{- 2 h}{h \left(x + 4\right) \left(x + h + 4\right)}$

Simplify

$f ' \left(x\right) = \lim h \to 0 \frac{- 2 \cancel{h}}{\cancel{h} \left(x + 4\right) \left(x + h + 4\right)}$

$f ' \left(x\right) = \lim h \to 0 \frac{- 2}{\left(x + 4\right) \left(x + h + 4\right)}$

Now we can substitute in a 0 for h

$f ' \left(x\right) = \frac{- 2}{\left(x + 4\right) \left(x + 0 + 4\right)}$

Simplify

$f ' \left(x\right) = \frac{- 2}{\left(x + 4\right) \left(x + 4\right)}$

Simplify

$f ' \left(x\right) = \frac{- 2}{{\left(x + 4\right)}^{2}}$

Watch this tutorial to see a similar question solved used the same methods.