Questions: Find the derivative of the function f(x) = 4 / (x - 1) using the limit definition of the derivative.

Solution Steps

To find the derivative of the function \( f(x) = \frac{4}{x-1} \) using the limit definition of the derivative, we will apply the formula for the derivative:

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

Substitute \( f(x) \) into this formula and simplify the expression to find the limit as \( h \) approaches 0.

To find the derivative of the function \( f(x) = \frac{4}{x-1} \) using the limit definition, we start by applying the formula for the derivative:

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

Substitute \( f(x) = \frac{4}{x-1} \) and \( f(x+h) = \frac{4}{x+h-1} \) into the formula:

\[ f'(x) = \lim_{h \to 0} \frac{\frac{4}{x+h-1} - \frac{4}{x-1}}{h} \]

Simplify the expression inside the limit:

\[ \frac{\frac{4}{x+h-1} - \frac{4}{x-1}}{h} = \frac{4(x-1) - 4(x+h-1)}{h(x+h-1)(x-1)} \]

This simplifies to:

\[ \frac{-4h}{h(x+h-1)(x-1)} \]

Cancel \( h \) in the numerator and denominator:

\[ \frac{-4}{(x+h-1)(x-1)} \]

Now, evaluate the limit as \( h \to 0 \):

\[ f'(x) = \lim_{h \to 0} \frac{-4}{(x+h-1)(x-1)} = \frac{-4}{(x-1)^2} \]

Final Answer