How do you find the derivative of #f(x)=1/x^2# using the limit process?

Question

Evelyn Agard · Accepted Answer

# f'(x)=-2/x^3 # By definition of the derivative # f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h # So with # f(x) = 1/x^2 # we have; # f'(x)=lim_(h rarr 0) ( 1/(x+h)^2 -1/x^2 ) / h # # :. f'(x)=lim_(h rarr 0)1/h*(x^2-(x+h)^2)/((x+h)^2x^2) # # :. f'(x)=lim_(h rarr 0) (x^2-(x^2+2hx+h^2))/(h(x+h)^2x^2) # # :. f'(x)=lim_(h rarr 0) (x^2-x^2-2hx-h^2)/(h(x+h)^2x^2) # # :. f'(x)=lim_(h rarr 0) (-2x-h)/((x+h)^2x^2) # # :. f'(x)=(-2x)/((x)^2x^2) # # :. f'(x)=-2/x^3 #

Cameron Alexander · Answer

undefined To find the derivative of $ f(x) = \frac{1}{x^2} $ using the limit process, you can follow these steps:

1. Start with the 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^2} $ into the definition:
$$ f'(x) = \lim_{h \to 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h} $$

3. Simplify the expression inside the limit:
$$ f'(x) = \lim_{h \to 0} \frac{x^2 - (x+h)^2}{h(x^2)(x+h)^2} $$

4. Expand $ (x+h)^2 $ and simplify the expression:
$$ f'(x) = \lim_{h \to 0} \frac{x^2 - (x^2 + 2hx + h^2)}{h(x^2)(x^2 + 2hx + h^2)} $$
$$ f'(x) = \lim_{h \to 0} \frac{x^2 - x^2 - 2hx - h^2}{h(x^2)(x^2 + 2hx + h^2)} $$
$$ f'(x) = \lim_{h \to 0} \frac{-2hx - h^2}{h(x^2)(x^2 + 2hx + h^2)} $$
$$ f'(x) = \lim_{h \to 0} \frac{-2x - h}{(x^2)(x^2 + 2hx + h^2)} $$

5. Cancel out the common factor of $ h $ in the numerator and denominator:
$$ f'(x) = \lim_{h \to 0} \frac{-2x - h}{(x^2)(x^2 + 2hx + h^2)} $$

6. Evaluate the limit as $ h $ approaches 0:
$$ f'(x) = \frac{-2x}{x^4} $$
$$ f'(x) = -\frac{2}{x^3} $$

So, the derivative of $ f(x) = \frac{1}{x^2} $ with respect to $ x $ is $ f'(x) = -\frac{2}{x^3} $.