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I've always learned implicit differentiation as $\frac{dy}{dx}$. But some tutorials on the web use $\frac{d}{dx}$. Here's an example of how I solve questions.

Find the slope of the tangent line to the graph of $x^2 + 3xy - 2y^2 = -4$ with respect to $x$.

I was taught that the derivative of $y$ with respect to $x$ is $\frac{dy}{dx}$. So I would do as followed:


  1. Find the derivative of each term using the product and power rule

    $(2x) + (3x \cdot \frac{dy}{dx} + 3y) + (- 4y \cdot \frac{dy}{dx}) = 0$


  1. I would then simplify the equation by moving all the terms containing $\frac{dy}{dx}$ to one side

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


  1. I would the factor out $\frac{dy}{dx}$ and simplify

    $\frac{dy}{dx}$(3x - 4y) = -2x - 3y$

    $\frac{dy}{dx} = \frac{-2x - 3y}{3x - 4y}$


So that's how I would find the derivative of an equation using implicit differentiation. But I checked the tutorial out on Khan Academy and they did it in a different way.

$x^2 + y^2 = 1$


$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = 0$

$2x + 2y \cdot \frac{dy}{dx} = 0$

$\frac{dy}{dx} = -\frac{x}{y}$

So I'm wondering, which way is better, and which one should I use? I believe that both the ways can be used on the same question, but I could be wrong. Any help would be highly appreciated, thanks!

  • They are doing it the same way. $\frac{d}{dx}$ is the operator (a function whose domain is functions and whose image is functions): you plug in a function, you get the derivative. So $\frac{d}{dx}(x^2)$ means “the derivative of $x^2$” and $\frac{d}{dx}(y^2)$ means “the derivative of $y^2$. It’s the same thing. They are exactly the same thing. Neither is “better” or “worse”, because they are the same. – Arturo Magidin Oct 20 '19 at 19:10
  • It is even better to say "I will denote the derivative with '." In this way, calculations are even cleaner: $(x^2+y^2)'=2x+2yy'.$ – William M. Oct 20 '19 at 19:12
  • Notation in mathematics is often a fine line between clarity, convenience & convention. Unfortunately the latter two elements can be trip wires while learning. – copper.hat Oct 20 '19 at 19:24
  • Is there a question that explains the latter method? –  Oct 20 '19 at 19:47
  • You should apply the two methods to the same problem to get a meaningful comparison of them. – amd Oct 20 '19 at 19:57
  • @amd Like I said before, I don't understand the latter question. So I was just using example problems. –  Oct 20 '19 at 23:45

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