Question

Derivadas. Utilize a definição f ' (x) = lim h-0 onde f(x+h) - f(x) para encontrar a derivada:
h

f ' (5) em que f(x) = 1 / 2x

Answer 1

$f\left(x \right )=\dfrac{1}{2x}\\ \\ \\ f'\left(x \right )=\lim_{h \to 0}\dfrac{f\left(x+h \right )-f\left(x \right )}{h}\\ \\ f'\left(x \right )=\lim_{h \to 0}\dfrac{\frac{1}{2\left(x+h \right )}-\frac{1}{2x}}{h}\\ \\ f'\left(x \right )=\lim_{h \to 0}\dfrac{\frac{x}{2x\left(x+h \right )}-\frac{x+h}{2x\left(x+h \right )}}{h}\\ \\ f'\left(x \right )=\lim_{h \to 0}\dfrac{\left(\frac{x-\left(x+h \right )}{2x\left(x+h \right )} \right )}{h}\\ \\ f'\left(x \right )=\lim_{h \to 0}\dfrac{\left(\frac{-h}{2x\left(x+h \right )} \right )}{h}\\ \\ f'\left(x \right )=\lim_{h \to 0}\left(\dfrac{-h}{2x\left(x+h \right )}\cdot \dfrac{1}{h} \right )\\ \\ f'\left(x \right )=\lim_{h \to 0}\dfrac{-1}{2x\left(x+h \right )}\\ \\ f'\left(x \right )=\dfrac{-1}{2x\left(x+0 \right )}\\ \\ \boxed{f'\left(x \right )=-\dfrac{1}{2x^{2}}}\\ \\ \\ f'\left(5 \right )=-\dfrac{1}{2\cdot 5^{2}}\\ \\ f'\left(x \right )=-\dfrac{1}{2 \cdot 25} \\ \\ \boxed{f'\left(x \right )=-\dfrac{1}{50}}$