# Implicit Confusion

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?

• yes.....and now? Commented Nov 26, 2016 at 0:26
• That is correct Commented Nov 26, 2016 at 0:27
• What is your question? Commented Nov 26, 2016 at 0:28
• @AlgorithmsX Are all my answers correct? Commented Nov 26, 2016 at 0:28

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 $y=\frac{3}{4}x$.
$$\frac{3x-2\cdot\frac{3}{4}x}{2x}=\frac{3x-\frac{3}{2}x}{2x}=\frac{\frac{3}{2}x}{2x}=\frac{3}{4}$$
• These ones were easy to show the substitution works. In a lot of case the substitute to change from one answer to another is much harder to find. In those problems to convince yourself it is the same select a point on your curve and substitute in the $(x,y)$ into each version of your derivative. Commented Nov 26, 2016 at 1:15