Course:MATH102/Question Challenge/2010 December 03

Question

Use the definition of the derivative (no differentiation rules) to find $f^{\prime }(2)$, where $f(x)=\frac{1}{x}$.

Solutions

Solution

$f(x)=\frac{1}{x}$ $f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$ $f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{\frac{1}{x+h}-\frac{-1}{x}}{h}$ $f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{\frac{x}{x(x+h)}-\frac{x+h}{x}}{h}$ $f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{\frac{-h}{x(x+h)}}{h}$ $f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{-h}{x(x+h)h}$ $f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{-1}{x(x+h)}$ $f^{\prime }(x)=\frac{-1}{x(x+(0))}$ $f^{\prime }(x)=\frac{-1}{x^2}$ Plugging in 2 we get, -1/4