How do you find f'(x) using the limit definition given $(1/x^2)$ ?

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Answer 1 · 2016-07-05T11:34:52

$= - 2/x^3$

$f(x) = 1/x^2$

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

$= lim_{h to 0} 1/h * (1/(x+h)^2 - 1/x^2)$

combining fractions:

$= lim_{h to 0} 1/h * (x^2- (x+h)^2)/(x^2(x+h)^2)$

$= lim_{h to 0} 1/h * (x^2- (x^2+2hx + h^2))/(x^2(x+h)^2$

$= lim_{h to 0} 1/h * (-2hx - h^2)/(x^2(x+h)^2$

$= lim_{h to 0} (-2x - h)/(x^2(x+h)^2$

$= lim_{h to 0} (-2x )/(x^2(x+h)^2) + mathcal(O)(h)$

$= lim_{h to 0} (-2 )/(x(x+h)^2) + mathcal(O)(h)$

$= (-2 )/(x(x)^2)$

$= - 2/x^3$

How do you find f'(x) using the limit definition given (1/x^2) (1/x^2) ?