Using the limit definition, how do you differentiate f(x)=4−2x−x2?

Question

belamontern9i · Accepted Answer

The limit definition of the derivative states that:

f^{'} (x) = lim_{△ x \to 0} \frac{f (x + △ x) - f (x)}{△ x}

For

f (x) = 2 - 2 x - x^{2}

we have:

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

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

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

f^{'} (x) = lim_{△ x \to 0} (- 2 - 2 x + △ x) = - 2 - 2 x

bondekoa6i · Accepted Answer

By definition:ddx(4−2x−x2)=limh→0(4−2(x+h)−(x+h)2)−(4−2x−x2)hddx(4−2x−x2)=limh→0(4−2x−2h−x2−2hx−h2−4+2x+x2)hddx(4−2x−x2)=limh→0−2h−2hx−h2hddx(4−2x−x2)=limh→0(−2−2x−h)ddx(4−2x−x2)=−2−2x

Using the limit definition, how do you differentiate f(x)=4-2x-x^{2}?

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