How to use fraction rules for dividing

Question

Hi I am very confused with fractions such as $\dfrac {\dfrac{a}{b}}{c}$ can this be simplified to $\dfrac {a}{b} \cdot \dfrac {1}{c}$?

so for example I was using the definition of a derivative to find the derivative of $\dfrac {9}{x}$, and I got to the point where my equation was $\dfrac {-9h}{x^2}+\dfrac {xh}{h}$. what I did was change it to $(-9h/x^2+xh)*(1/h)$ and I could cancel the $h$ from $-9h$ and $1/h$ to get the right answer of $-9/x^2$.

But when I simplify $\dfrac {\sin x}{\tan x}$ using this I get

 sin x/sin x/cos x 
= sinx/sinx * 1/cosx 
= 1/cosx 

but this is wrong. the correct answer is $\cos x$.

so why does this approach work for some questions but not others?

Answer 1

Your mistake for the second fraction was taking the fraction $\dfrac {\sin x}{\dfrac{\sin x}{\cos x}}$ with $\dfrac {\sin x}{\sin x}$ as the numerator, but when you work with compound fractions like these, the numerator is actually $\dfrac {\sin x}{1}$ and the denominator is $\tan x = \dfrac {\sin x}{\cos x}$, to wit...

$$\dfrac {\sin x}{\tan x} = \left(\dfrac {\dfrac {\sin x}{1}}{\dfrac{\sin x}{\cos x}}\right)=\sin x\cdot {\dfrac {\cos x}{\sin x}} = \cos x$$

Answer 2

I think the confusion your having is that $a/b/c$ is an ambiguous expression, as noted by the other commenters. It can both be represented as, $\tfrac{(\tfrac{a}{b})}{c}$ or $\tfrac{a}{(\tfrac{b}{c})}$

If its the first one, we multiply by $\tfrac{\tfrac{1}{c}}{\tfrac{1}{c}}$ to get $\tfrac{(\tfrac{a}{b})}{c} \cdot \tfrac{\tfrac{1}{c}}{\tfrac{1}{c}} = \tfrac{\tfrac{a}{b} \cdot \tfrac{1}{c}}{c \cdot\tfrac{1}{c}} = \tfrac{\tfrac{a}{b} \cdot \tfrac{1}{c}}{1} = \tfrac{a}{b} \cdot \tfrac{1}{c}$

This is not necessarily the case for the second interpretation of the expression.