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

Question

Paisley Johnson · Accepted Answer

#-1/x^2#

Differentiate from #color(blue)"first principles"#

#color(red)(|bar(ul(color(white)(a/a)color(black)(f'(x)=lim_(hto0)(f(x+h)-f(x))/h)color(white)(a/a)|)))#

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.

#f'(x)=lim_(hto0)(1/(x+h)-1/x)/h#

combine numerator into a single fraction

#=lim_(hto0)(1/(x+h) xxx/x-1/x xx(x+h)/(x+h))/h#

#=lim_(hto0)(x/(x(x+h))-(x+h)/(x(x+h)))/h=lim_(hto0)(cancel(x)-cancel(x)-h)/(hx(x+h))#

We can now 'cancel' h

#=lim_(hto0)(-cancel(h)^1)/(cancel(h)^1x(x+h))#

as #hto0rArrf'(x)=-1/x^2#

Adam Adkins · Answer

undefined 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} $.

William Abdalla · Answer

undefined 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} $.