"How do you use the definition of a..." was answered by our experts

Hallie Watts

Answered question

2022-04-11

How do you use the definition of a derivative to find the derivative of

\frac{4}{x^{2}}

Answer & Explanation

tabuevniru8op

Beginner2022-04-12Added 14 answers

Just remember that

\frac{4}{x^{2}} = 4 x^{- 2}

and that

{(x^{n})}^{'} = n x^{n - 1}

.
So

{(\frac{4}{x^{2}})}^{'} = {(4 x^{- 2})}^{'} = 4 \times (- 2) x^{- 3} = - \frac{8}{x^{3}}

YAMAGAMA699k

Beginner2022-04-13Added 8 answers

Let $f (x) = \frac{4}{x^{2}}$
I will assume that you are using the limit definition in the form
$f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$
So, we get:
$f^{'} (x) = lim_{h \to 0} \frac{\frac{4}{{(x + h)}^{2}} - \frac{4}{x^{2}}}{h}$ (indeterminate form $\frac{0}{0}$ )
$= lim_{h \to 0} \frac{\frac{4 x^{2} - (4 {(x + h)}^{2})}{x^{2} ({(x + h)}^{2})}}{\frac{h}{1}}$
$= lim_{h \to 0} \frac{4 x^{2} - (4 x^{2} + 8 x h + 4 h^{2})}{x^{2} {(x + h)}^{2}} \cdot \frac{1}{h}$
$= lim_{h \to 0} \frac{- 8 x h - 4 h^{2}}{x^{2} {(x + h)}^{2}} \cdot \frac{1}{h}$
$= lim_{h \to 0} \frac{- 8 x - 4 h}{x^{2} {(x + h)}^{2}}$ (not an indeterminate form)
$= \frac{- 8 x}{x^{4}} = \frac{- 8}{x^{3}}$
If you are using $f^{'} (x) = lim_{t \to x} \frac{f (x) - f (t)}{x - t}$
Then the algebra is similar.
$\frac{f (x) - f (t)}{x - t} = \frac{\frac{4}{x^{2} - \frac{4}{t^{2}}}}{x - t}$
$= \frac{\frac{4 t^{2} - 4 x^{2}}{x^{2} t^{2}}}{\frac{x - t}{1}}$
$= \frac{4 (t^{2} - x^{2})}{x^{2} t^{2}} \cdot \frac{1}{x - t}$
$= \frac{- 4 (x^{2} - t^{2})}{x^{2} t^{2}} \cdot \frac{1}{x - t}$
$= \frac{- 4 (x + t) (x - t)}{x^{2} t^{2}} \cdot \frac{1}{x - t}$
$= \frac{- 4 (x + t)}{x^{2} t^{2}}$
Taking the limit as $t \to x$ , we get
$\frac{- 4 (2 x)}{x^{2} x^{2}} = \frac{- 8}{x^{3}}$

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