Question

por favor me ajudem!
Calcular f'(x) pela definição:
f(x)=1/x²

Answer 1

A letra pode estar meio feia, mas qualquer dúvida, só chamar

Answer 2

Definição de derivada

$\large\boxed{\boxed{\displaystyle\mathsf{f'(x)=\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}}}}$

$\displaystyle\mathsf{f'(x)=\lim_{h\to 0}\dfrac{\frac{1}{{(x+h)}^{2}}-\frac{1}{{x}^{2}}}{h}}$

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

$\displaystyle\mathsf{f'(x)=\lim_{h \to 0}\dfrac{1}{h}\left(\dfrac{{x}^{2}-{(x+h)}^{2}}{{x}^{2}.{(x+h)}^{2}}\right)}$

$\displaystyle\mathsf{f'(x)=\lim_{h \to 0}\dfrac{1}{h}\left(\dfrac{{x}^{2}-{x}^{2}-2hx - {h}^{2}}{{x}^{2}.{(x+h)}^{2}}\right)}$

$\displaystyle\mathsf{f'(x)=\lim_{h \to 0}\dfrac{1}{h}\left(\dfrac{-2hx-{h}^{2}}{{x}^{2}.{(x+h)}^{2}}\right)}$

$\displaystyle\mathsf{f'(x)=\lim_{h \to 0}\dfrac{1}{\cancel{h}}\left(\dfrac{\cancel{h}(-2x-h) }{{x}^{2}.{(x+h)}^{2}}\right)}$

$\mathsf{f'(x)=\dfrac{-2x-0}{{x}^{2}.{(x+0)}^{2}} =-\dfrac{2{x}}{ {x}^{4}} = - \dfrac{2}{ {x}^{3} } }$

$\huge\boxed{\boxed{\boxed{\mathsf{f'(x)=-\dfrac{2}{{x}^{3}}}}}}$