How do you use the limit definition of the derivative to find the derivative of #f(x)=1/x#?

The aim is to eliminate the h on the denominator otherwise division by zero which is undefined. Manipulate the numerator to obtain h as a factor, to cancel with h on denominator.

combine numerator into a single fraction

We can now 'cancel' h

To find the derivative of $f(x) = \frac{1}{x}$ using the limit definition of the derivative:

1. Start with the definition of the derivative: $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$.

2. Substitute $f(x) = \frac{1}{x}$ into the definition.

3. Simplify the expression: $f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h}$.

4. Combine the fractions: $f'(x) = \lim_{h \to 0} \frac{x - (x + h)}{hx(x + h)}$.

5. Simplify the numerator: $f'(x) = \lim_{h \to 0} \frac{-h}{hx(x + h)}$.

6. Cancel out $h$ from the numerator and denominator: $f'(x) = \lim_{h \to 0} \frac{-1}{x(x + h)}$.

7. Evaluate the limit as $h$ approaches 0: $f'(x) = \frac{-1}{x^2}$.

So, the derivative of $f(x) = \frac{1}{x}$ is $f'(x) = -\frac{1}{x^2}$.

To use the limit definition of the derivative to find the derivative of $f(x) = \frac{1}{x}$, follow these steps:

1. Start with the limit definition of the derivative:

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

2. Substitute the function $f(x) = \frac{1}{x}$ into the limit definition:

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

3. Combine the fractions in the numerator:

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

4. Simplify the expression in the numerator:

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

5. Cancel out the $h$ terms:

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

6. Now, take the limit as $h$ approaches 0:

$f'(x) = \frac{-1}{x^2}$

So, the derivative of $f(x) = \frac{1}{x}$ with respect to $x$ is $f'(x) = \frac{-1}{x^2}$.