Given $h(x) = \frac{f(x)}{g(x)}$, find $h'(3)$
$f(1)=2, f'(1)=-1, g(1)=1, g'(1)=2 $
$f(2)=1, f'(2)=0.5, g(2)=3, g'(2)=0 $
$f(3)=3, f'(3)=2, g(3)=1, g'(3)=-2$
What I am getting is $-7/2$ , but I am unsure if that is true because I believe the correct answer is $-3/4$. Basically I found the derivative of $f(x)\times g(x)^{-1}$, and then plugged in the appropriate values from the list above.
By the quotient rule, $$h'(x) = \frac{g(x)f'(x) - f(x)g'(x)}{g(x)^2},$$ so that $$h'(3) = \frac{g(3)f'(3) - f(3)g'(3)}{g(3)^2} = \frac{2-(-3)}{1^2} = 5.$$