Using the limit definition, how do you find the derivative of #f(x)=3(x^(-2)) #?

Question

Evelyn Agard · Accepted Answer

#f'(x) = -6x^(-3)#

According to the limit definition, the derivative of #f(x)# is

#f'(x) = lim_(h -> 0) (f(x+h) - f(x)) / h#

In your case, this means

#f'(x) = lim_(h -> 0) (3(x+h)^(-2) - 3x^(-2)) /h#

# = lim_(h -> 0) (3/(x+h)^2 - 3/x^2) /h#

# = lim_(h -> 0) ((3x^2 - 3(x+h)^2)/((x+h)^2 *x^2)) /h#

# = lim_(h -> 0) (3x^2 - 3(x+h)^2)/(h(x+h)^2 *x^2) #

... use #(a+b)^2 = a^2 + 2ab + b^2#...

# = lim_(h -> 0) (3x^2 - 3(x^2 + 2xh + h^2))/(h(x+h)^2 *x^2) #

# = lim_(h -> 0) (cancel(3x^2) - cancel(3x^2) - 6xh - 3h^2)/(h(x+h)^2 *x^2) #

# = lim_(h -> 0) (h( - 6x - 3h))/(h(x+h)^2 *x^2) #

... cancel #h#....

# = lim_(h -> 0) ( - 6x - 3h)/((x+h)^2 *x^2) #

... at this point you can apply the #lim#, so plug #h=0#:

# = ( - 6x - 0)/((x+0)^2 *x^2) #

# = ( - 6x )/(x^4) #

# = ( - 6 )/(x^3) #

# = -6x^(-3) #

Thus, your derivative is

#f'(x) = -6x^(-3)#

Axel Alvarez · Answer

undefined To find the derivative of $ f(x) = 3x^{-2} $ using the limit definition, we start by using the definition of the derivative:

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

Substitute the given function $ f(x) = 3x^{-2} $ into the definition:

$$ f'(x) = \lim_{h \to 0} \frac{3(x + h)^{-2} - 3x^{-2}}{h} $$

Expand and simplify the expression:

$$ f'(x) = \lim_{h \to 0} \frac{3}{h} \left( \frac{1}{(x + h)^2} - \frac{1}{x^2} \right) $$

$$ f'(x) = \lim_{h \to 0} \frac{3}{h} \left( \frac{x^2 - (x + h)^2}{(x^2)((x + h)^2)} \right) $$

$$ f'(x) = \lim_{h \to 0} \frac{3}{h} \left( \frac{x^2 - (x^2 + 2xh + h^2)}{(x^2)((x + h)^2)} \right) $$

$$ f'(x) = \lim_{h \to 0} \frac{3}{h} \left( \frac{-2xh - h^2}{(x^2)((x + h)^2)} \right) $$

$$ f'(x) = \lim_{h \to 0} \frac{-6x - 3h}{(x^2)((x + h)^2)} $$

Now, as $ h $ approaches 0, the expression becomes:

$$ f'(x) = \frac{-6x}{(x^2)(x^2)} $$

$$ f'(x) = \frac{-6}{x^3} $$

So, the derivative of $ f(x) = 3x^{-2} $ is $ f'(x) = \frac{-6}{x^3} $.