Quiz 5 (Solution)


Date: MATH 1210-4 - Spring 2004

  1. Use the definition of derivative, $ f'(x) = \lim_{h\rightarrow 0}
\frac{f(x+h) -f(x)}{h}$, to find the derivative of $ f(x) =
\frac{1}{x^2+1}$.

    \begin{displaymath}\begin{split}f'(x) &= \lim_{h\rightarrow 0} \frac{f(x+h) -f(x...
...rrow 0}(x^2+1)((x+h)^2+1)} = \frac{-2x}{(x^2+1)^2}. \end{split}\end{displaymath}    

    Thus, $ f'(x) = \frac{-2x}{(x^2+1)^2}$.

  2. The limit $ \lim_{h\rightarrow 0} \frac{(2+h)^5-3(2+h) - 26}{h}$ is a derivative $ f'(a)$ of a function $ f(x)$ at a point $ a$. Find $ f(x)$ and the point $ a$.

    We have

    $\displaystyle \lim_{h\rightarrow 0} \frac{(2+h)^5-3(2+h) - 26}{h} = \lim_{h\rightarrow 0} \frac{f(2+h)-f(2)}{h}$    

    for $ f(x) = x^5-3x$, (check: $ f(2+h)-f(2) = (2+h)^5-3(2+h) - (2^5 -
3\cdot 2) = (2+h)^5-3(2+h) -26$) so that the given limit is the derivative of $ x^5-3x$ at $ a=2$.



Javier Fernandez 2004-03-01