Derivadas: Problemas y más.

Colección de 1 ejercicio resueltos relacionados con las derivadas.

Utilizando la definición de derivada, calcula $f^{\prime }(-2)$, siendo $f(x)={\displaystyle \frac{1}{x}}$:

$$f^{\prime }(-2)={\displaystyle \underset{h\to 0}{{\displaystyle \mathit{\text{l}}\acute{ı}\mathit{\text{m}}}}{\displaystyle \frac{f(-2+h)-f(-2)}{h}}={\displaystyle \underset{h\to 0}{{\displaystyle \mathit{\text{l}}\acute{ı}\mathit{\text{m}}}}{\displaystyle \frac{{\displaystyle \frac{1}{-2+h}}-{\displaystyle \frac{1}{-2}}}{h}}={\displaystyle \underset{h\to 0}{{\displaystyle \mathit{\text{l}}\acute{ı}\mathit{\text{m}}}}{\displaystyle \frac{{\displaystyle \frac{1}{-2+h}}+{\displaystyle \frac{1}{2}}}{h}}={\displaystyle \underset{h\to 0}{{\displaystyle \mathit{\text{l}}\acute{ı}\mathit{\text{m}}}}{\displaystyle \frac{{\displaystyle \frac{2-2+h}{-4+2h}}}{h}}=}}}}$$

$$={\displaystyle \underset{h\to 0}{{\displaystyle \mathit{\text{l}}\acute{ı}\mathit{\text{m}}}}{\displaystyle \frac{h}{(-4+2h)\cdot h}}=\left\{{\displaystyle \frac{0}{0}}\right\}={\displaystyle \underset{h\to 0}{{\displaystyle \mathit{\text{l}}\acute{ı}\mathit{\text{m}}}}{\displaystyle \frac{\cancel{h}}{(-4+2h)\cdot \cancel{h}}}={\displaystyle \underset{h\to 0}{{\displaystyle \mathit{\text{l}}\acute{ı}\mathit{\text{m}}}}{\displaystyle \frac{1}{-4+2h}}={\displaystyle \frac{-1}{4}}}}}$$

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