Implicit Differentiation Doubt

Question

One of my classmates said that for $x^2+y^2=1$, to find $\frac{dy}{dx}$, the following method can be used:

First rearrange the equation, $$x^2+y^2-1=0$$. Then assume $$w=x^2+y^2-1$$ $$\frac{dy}{dx}=\frac{dw}{dx} \div \frac{dw}{dy}$$

Also, when finding $\frac{dw}{dx}$, $y$ is considered a constant. And similarly, when finding $\frac{dw}{dy}$, $x$ is considered a constant.

Let me proceed. $\frac{dw}{dx}=2x$ regarding $y$ as a constant. $\frac{dw}{dy}=2y$ regarding $x$ as a constant. $\frac{dy}{dx}=2x/2y$ which is the negative of the correct answer.

It seems like it applies for all the equations.

I think his methods is completely unreasonable but by using his method the result is always the negative of the correct result. Why????

Answer 1

What he is showing you is the result of partial differentiation. In partial differentiation, for a function $$u = f(x, y),$$ then $$du = \partial_x u + \partial_y u$$ where $\partial_x u$ is the partial differential of $u$ with respect to $x$ (i.e., differentiating with all variables but $x$ held constant).

Since, in your case, $u$ (and therefore $du$) is zero, we can then say:

$$ 0 = \partial_x u + \partial_y u $$

Now, if we solve for one of them, we get:

$$ \partial_x u = -\partial_y u $$

Or, in other words,

$$ \frac{\partial_x u}{\partial_y u} = -1 $$

That's for differentials. The partial derivative of $u$ with respect to $x$ is actually $\frac{\partial_x u}{dx}$. Now, let's multiply both sides by $\frac{dy}{dx}$:

$$\frac{\partial_x u}{\partial_y u} \frac{dy}{dx} = -\frac{dy}{dx} $$

We can rearrange the left-hand side so that we get partial derivatives:

$$\frac{\frac{\partial_x u}{dx}}{\frac{\partial_y u}{dy}} = -\frac{dy}{dx} $$

And that is the procedure that your friend gave you. However, for a general technique for implicit differentiation, the method that @MrYouMath above gives is generally more straightforward.

Answer 2

You do not have to rearrange the equation just apply the total differential to the equation

$$x^2+y^2=1 \implies d(x^2)+d(y^2)=d(1) \implies 2xdx+2ydy = 0.$$

Now, solve for $dy/dx$.

Answer 3

Similar to the other answers, I would differentiate with respect to $x$ and treat $y=y(x)$ by computing \begin{align} \frac{d}{dx}\left(x^2+y(x)^2\right) &= \frac{d}{dx}(1)\\ \implies 2x+2y\frac{dy}{dx} &= 0\\ \implies \frac{dy}{dx} &= -\frac{x}{y} \qquad \blacksquare \end{align}

Answer 4

It is quite reasonable. Note: $$x^2+y^2=1 \Rightarrow w(x,y)=x^2+y^2-1=0.$$ The total differential of the function $w(x,y)$ is: $$w_xdx+w_ydy=0 \Rightarrow \frac{dy}{dx}=-\frac{w_x}{w_y}=-\frac{\frac{dw}{dx}}{\frac{dw}{dy}}.$$ So your classmate is only missing minus.