Implicit Confusion

Question

If I solved for $y$ and then calculated the derivative of:

$$\frac{3x^2}{4y}=x$$

I would get $\frac{3}{4}$ right? Because if I solved for $y$ I would get $\frac{3}{4}x=y$ and then $\frac{dy}{dx}=\frac{3}{4}.$ Using three different methods of differentiation I got $\frac{16y^2-24xy}{-12x^2}$ by first applying quotient rule, and then solving for $\frac{dy}{dx}$; my second method was multiplying both sides of the equation by the denominator of the left side, and then using implicit differentiation I got $\frac{3x-2y}{2x}=\frac{dy}{dx}$. Are all these methods correct, since there are three answers wouldn't they be equal to each other?

Answer 1

All of these are correct. What you need to remember is that $y$ and $x$ and linked by the fact that $y=\frac{3}{4}x$. So any answer can be reached from another via substitution.

Take $\frac{16y^2-24xy}{-12x^2}$ and substitute in $y=\frac{3}{4}x$.

$$\frac{16\left(\frac{3}{4}x\right)^2-24x\left(\frac{3}{4}x\right)}{-12x^2}=\frac{9x^2-18x^2}{-12x^2}=\frac{-9x^2}{-12x^2}=\frac{3}{4}$$

Or take $\frac{3x-2y}{2x}$ and substitute in $y=\frac{3}{4}x$.

$$\frac{3x-2\cdot\frac{3}{4}x}{2x}=\frac{3x-\frac{3}{2}x}{2x}=\frac{\frac{3}{2}x}{2x}=\frac{3}{4}$$