Question

calcule as derivadas das expressões abaixo, usando as fórmulas de derivação;
$f(x) = \frac{2}{ {x}^{2} }$
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Answer 1

Resposta:

$f'(x)=-\frac{4}{{x}^{3}}$

Explicação passo-a-passo:

$f(x) = \frac{2}{ {x}^{2} } \\ f(x) = \frac{g(x)}{h(x)} \\ \\ f'(x) = \frac{g'(x).h(x) - g(x).h'(x)}{ {(h(x))}^{2} } \\ g'(x) = 0 \\ h'(x) = 2x \\ \\ f'(x) = \frac{0.{x}^{2} - 2.2x }{ {( {x}^{2} )}^{2} } = -\frac{4x}{ {x}^{4} } = -\frac{4}{ {x}^{3} }$

Answer 2

Resposta:

Explicação passo-a-passo:

$f'(x)= \lim_{h \to 0}\dfrac{f(x+h)-f(x)}{h}\\ \\f'(x)= \lim_{h \to 0}\dfrac{\frac{2}{(x+h)^{2}}-\frac{2}{x^{2}}}{h}\\\\f'(x)= \lim_{h \to 0}\dfrac{\frac{2x^{2}-2(x+h)^{2}}{(x+h)^{2}.x^{2}}}{h}\\\\f'(x)= \lim_{h \to 0}\dfrac{2x^{2}-2(x+h)^{2}}{h.(x+h)^{2}.x^{2}}}\\\\f'(x)= \lim_{h \to 0}\dfrac{2x^{2}-2(x^{2}+2xh+h^{2})}{h.(x+h)^{2}.x^{2}}}\\\\f'(x)= \lim_{h \to 0}\dfrac{2x^{2}-2x^{2}-4xh-2h^{2}}{h.(x+h)^{2}.x^{2}}}\\\\f'(x)= \lim_{h \to 0}\dfrac{h.(-4x-2h)}{h.(x+h)^{2}.x^{2}}}$

$f'(x)= \lim_{h \to 0}\dfrac{(-4x-2h)}{(x+h)^{2}.x^{2}}}\\\\\\f'(x)= \dfrac{-4x-2.0}{(x+0)^{2}.x^{2}}}\\\\\\f'(x)= \dfrac{-4x}{x^{2}.x^{2}}}\\\\\\f'(x)= \dfrac{-4x}{x^{4}}}\\\\\\f'(x)= -\dfrac{4}{x^{3}}}$