# Question #65e7a

Nov 6, 2016

$\frac{- 1}{\left(x + h - 8\right) \left(x - 8\right)}$

#### Explanation:

GIven: $f \left(x\right) = \frac{1}{x - 8}$

Then $f \left(x + h\right) = \frac{1}{x + h - 8}$

Substituting into the difference quotient:

$\frac{\frac{1}{x + h - 8} - \frac{1}{x - 8}}{h}$

Multiply by 1 in the form $\frac{\left(x + h - 8\right) \left(x - 8\right)}{\left(x + h - 8\right) \left(x - 8\right)}$

$\frac{\frac{1}{x + h - 8} - \frac{1}{x - 8}}{h} \frac{\left(x + h - 8\right) \left(x - 8\right)}{\left(x + h - 8\right) \left(x - 8\right)}$

Multiply by the 1s in the upper numerators and h in the lower denominator:

$\frac{\frac{\left(x + h - 8\right) \left(x - 8\right)}{x + h - 8} - \frac{\left(x + h - 8\right) \left(x - 8\right)}{x - 8}}{h \left(x + h - 8\right) \left(x - 8\right)}$

Now, I will show you which factors cancel in the upper part:

$\frac{\frac{\cancel{\left(x + h - 8\right)} \left(x - 8\right)}{\cancel{\left(x + h - 8\right)}} - \frac{\left(x + h - 8\right) \cancel{\left(x - 8\right)}}{\cancel{\left(x - 8\right)}}}{h \left(x + h - 8\right) \left(x - 8\right)}$

Remove the canceled factors:

$\frac{\left(x - 8\right) - \left(x + h - 8\right)}{h \left(x + h - 8\right) \left(x - 8\right)}$

Distribute the implicit -1:

$\frac{x - 8 - x - h + 8}{h \left(x + h - 8\right) \left(x - 8\right)}$

When we combine like terms in the numerator, we are left with -h:

$\frac{- h}{h \left(x + h - 8\right) \left(x - 8\right)}$

$- \frac{h}{h}$ reduces to -1:

$\frac{- 1}{\left(x + h - 8\right) \left(x - 8\right)}$

Now it will be safe for you to let $h \to 0$