Browse Questions

# Find the derivative of $99x$ at $x = 100$.

Let $f(x)=99x$. Accordingly,
$f'(100) \lim\limits_{h \to 0} \large\frac{f(100+h)-f(100)}{h}$
$= \lim\limits_{h \to 0} \large\frac{99(100+h)-99(100)}{h}$
$= \lim\limits_{h \to 0} \large\frac{99 \times 100+99h-99 \times 100}{h}$
$= \lim\limits_{h \to 0} \large\frac{99h}{h}$
$= \lim\limits_{h \to 0}(99) = 99$
Thus, the derivative of $99x$ at $x =100$ is $99$