se f(x)=2/x^3, calcular f´(-2)
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Primeiro, vamos derivar, pela regra do tombo:
![\large{\boxed{\bf(x) = x^n\rightarrow\ f'(x) = n~\cdot~x^{n-1}} \large{\boxed{\bf(x) = x^n\rightarrow\ f'(x) = n~\cdot~x^{n-1}}](https://tex.z-dn.net/?f=%5Clarge%7B%5Cboxed%7B%5Cbf%28x%29+%3D+x%5En%5Crightarrow%5C+f%27%28x%29+%3D+n%7E%5Ccdot%7Ex%5E%7Bn-1%7D%7D)
Então, temos que:
![f(x) = \frac{2}{x^3} = 2~\cdot~x^{-3}\\ \\ f'(x) = 2~\cdot-3\ x^{-3-1}\\ \\ f'(x) = -6x^{-4}\\ \\ f'(x) = \frac{-6}{x^4}\\ \\ f'(-2) = -6(-2)^{-4}\\ \\ f'(-2) = -6(\frac{1}{16})\\ \\ \large\boxed{ f'(-2) = -\frac{3}{8}} f(x) = \frac{2}{x^3} = 2~\cdot~x^{-3}\\ \\ f'(x) = 2~\cdot-3\ x^{-3-1}\\ \\ f'(x) = -6x^{-4}\\ \\ f'(x) = \frac{-6}{x^4}\\ \\ f'(-2) = -6(-2)^{-4}\\ \\ f'(-2) = -6(\frac{1}{16})\\ \\ \large\boxed{ f'(-2) = -\frac{3}{8}}](https://tex.z-dn.net/?f=f%28x%29+%3D+%5Cfrac%7B2%7D%7Bx%5E3%7D+%3D+2%7E%5Ccdot%7Ex%5E%7B-3%7D%5C%5C+%5C%5C+f%27%28x%29+%3D+2%7E%5Ccdot-3%5C+x%5E%7B-3-1%7D%5C%5C+%5C%5C+f%27%28x%29+%3D+-6x%5E%7B-4%7D%5C%5C+%5C%5C+f%27%28x%29+%3D+%5Cfrac%7B-6%7D%7Bx%5E4%7D%5C%5C+%5C%5C+f%27%28-2%29+%3D+-6%28-2%29%5E%7B-4%7D%5C%5C+%5C%5C+f%27%28-2%29+%3D+-6%28%5Cfrac%7B1%7D%7B16%7D%29%5C%5C+%5C%5C+%5Clarge%5Cboxed%7B+f%27%28-2%29+%3D+-%5Cfrac%7B3%7D%7B8%7D%7D)
Então, temos que:
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