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 $f(x) = \frac{1}{x^2+1}$.

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

   
   

   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

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

   
   
   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 $a=2$.