1. ## Quotient rule

Hi all,

I'm struggling to solve this equation using the quotient rule.

y = 3cos2x
2sin3x

Any help appreciated

2. Originally Posted by astrostar
Hi all,

I'm struggling to solve this equation using the quotient rule.
y = 3cos2x
2sin3x

Any help appreciated
If you can't remember the quotient rule, you can use the product rule.

$\displaystyle \frac d{dx} \frac {3cos(2x)}{2sin(3x)} ~~~~=~~~~ \frac 32 \frac d{dx} \frac 1{sin(3x)} cos(2x) ~~~~=~~~~ \frac 32 \frac d{dx} [sin(3x)]^{-1}* cos(2x)$

The quotient rule is just a version of the product rule, if you use the product rule on the version of your problem that I showed you, you will end up getting your problem into the same format as the quotient rule after you simplify.

3. ## If you mean

you are trying to find $\displaystyle f'(x)$ when $\displaystyle f(x)=\frac{3cos(2x)}{2sin(3x)}$...then here is what you do...its Low(D[hi])-hi[D(low)]/low˛...so $\displaystyle f'(x)=\frac{2sin(3x)[3\cdot{-sin(2x)}\cdot{2}]-3cos(2x)[2\cdot{cos(3x)}\cdot{3}]}{4sin(3x)^2}$...which simplifies to $\displaystyle f'(x)=\frac{-3(3cos(2x)\cdot{cos(3x)}+2sin(2x)\cdot{sin(3x)})}{ 2sin(3x)^2}$

4. Thanks for the replies,

I have to find the answer using the quotient rule, my workng is,

y = 3cos2x
2sin3x

let u = 3cos2x

du = -6sinx
dx

let v = 2sin3x

dv = 6cos3x
dx

this is where I get stuck

5. ## Ok your doing it a weird way

but if you want to do it that way youd say that $\displaystyle \frac{D[\frac{u}{v}]}{dx}=\frac{v\cdot\frac{du}{dx}-u\cdot\frac{dv}{dx}}{v^2}$