Question

Find the derivative of the function using the definition of derivative. f(x) = (8 + x )/(1 − 8x)

Answer 1

\[
f'(x) = \lim_{h \rightarrow 0}\frac{f(x+h) - f(x)}{h} ~~~~~~~ \text{----- (1)}\\
f(x) = \frac{8+x}{1-8x} \\
f(x+h) = ? \\
\text{ } \\
\text{Plug f(x+h) and f(x) into (1), simplify and take the limit}
\]

Answer 2

How do I simplify it? That is what I'm having difficulty with.

Answer 3

\[
f'(x) = \lim_{h \rightarrow 0}\frac{f(x+h) - f(x)}{h} ~~~~~~~ \text{----- (1)}\\
f(x) = \frac{8+x}{1-8x} \\
f(x+h) = \frac{8+x+h}{1-8x-8h} \\
f(x+h) - f(x) = \frac{8+x+h}{1-8x-8h} - \frac{8+x}{1-8x} = \\
\frac{(8+x+h)(1-8x)-(8+x)(1-8x-8h)}{(1-8x-8h)(1-8x)} = \\
\frac{(8+x+h)(1-8x)-(8+x)(1-8x-8h)}{(1-8x-8h)(1-8x)} = \\
\frac{\cancel{8}-\cancel{64x}+\cancel{x}-\cancel{8x^2}+h-\cancel{8xh}-\cancel{8}+\cancel{64x}+64h-\cancel{x}+\cancel{8x^2}+\cancel{8xh}}{(1-8x-8h)(1-8x)} = \\
\frac{65h}{(1-8x-8h)(1-8x)} \\
f'(x) = \lim_{h \rightarrow 0}\frac{f(x+h) - f(x)}{h} =
\lim_{h \rightarrow 0}\frac{65h}{h(1-8x-8h)(1-8x)} = \\
\lim_{h \rightarrow 0}\frac{65}{(1-8x-8h)(1-8x)} = \frac{65}{(1-8x)(1-8x)} =
\frac{65}{(1-8x)^2}
\]