How do you use the limit definition to find the derivative of $y=1/(x+2)$ ?

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Answer 1 · 2016-11-13T15:11:00

Use the formula $f'(x) = lim_(h->0) (f(x + h) - f(x))/h$ .

$f'(x) =lim_(h->0) (1/(x + h + 2) - 1/(x +2))/h$

$f'(x) = lim_(h-> 0) ((x + 2)/((x + 2)(x + h + 2)) - (x+ h + 2)/((x + 2)(x + h + 2)))/h$

$f'(x) = lim_(h->0) ((x + 2 - x - h - 2)/((x + 2)(x + h + 2)))/h$

$f'(x) = lim_(h->0) (-h)/((x+ 2)(x + h + 2)) xx 1/h$

$f'(x) = lim_(h->0) -1/((x + 2)(x + h + 2))$

We can now substitute directly.

$f'(x) = -1/((x+ 2)(x + 0 + 2))$

$f'(x) = -1/(x + 2)^2$

Verification with the quotient rule yields the same result:

Letting $f(x) = (g(x))/(h(x))$ with $g(x) = 1$ and $h(x) = x + 2$ , $f'(x)$ is given by $f'(x) = (g'(x) xx h(x) - g(x) xx h'(x))/(h(x))^2$ .

Calculating, we have:

$f'(x) = (0(x + 2) - 1(1))/(x + 2)^2$

$f'(x) = -1/(x + 2)^2$ , which is the answer we obtained previously.

Hopefully this helps!

How do you use the limit definition to find the derivative of y=1/(x+2)y=1/(x+2)?