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.
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 $\displaystyle f'(x)=\frac{-3(3cos(2x)\cdot{cos(3x)}+2sin(2x)\cdot{sin(3x)})}{ 2sin(3x)^2}$