Yesterday, this question posted within the AP Calculus Community about implicit derivatives.

Below, I argue why the derivatives MUST be the same, show how four different variations can all be shown to give the same derivative, and provide a final conclusion.

**INITIAL INTUITION**

The Desmos graph of the given relation, is , is shown below. Logically, it seems that even when the terms of the relation are algebraically rearranged, the graph should be invariant. The other two forms mentioned in the Community post are on lines 2 and 3. Lines 4, 5, and 6 show three other variations. Here is the link to my Desmos graph allowing you to change between the forms to visually confirm the graphical invariance intuition.

If calculus “works”, it also shouldn’t matter how one calculates a derivative. While the forms of the derivative certainly could LOOK different, because any point on the invariant graph has the same tangent line no matter what the form of its equation, and the derivative of a relation at a point is the slope of that invariant tangent line, then the derivative also MUST be invariant.

**CALCULATING “DIFFERENT” DERIVATIVES**

To show the derivatives are fundamentally all the same (as suspected by the initial post), I calculate the derivatives of the equations on lines 1 and 3 given in the initial post as well as my variations on lines 4 and 6.

LINE 1:

Using the Chain Rule on the left and the Quotient Rule on the right gives

LINE 3:

This version is more complicated, requiring the Product Rule in addition to the earlier Chain and Quotient Rules. In the penultimate line, I used the original equation to substitute for to transform the derivative into the same form as line 1.

LINE 4:

This time, differentiation requires only the Chain and Product Rules.

After the usual substitution for , I multiplied both sides by to clear the denominator and solved for , returning the same result.

LINE 6:

This time, the relation is solved for x, resulting in a much more complicated Quotient+Chain Rule calculation, but substituting for and changing the form leads once again to the same answer.

Hopefully this is convincing evidence that all derivative forms can be shown to be equivalent. If you’re still learning implicit differentiation, I encourage you to show the derivatives from the lines 2 and 5 variations are also equivalent.

**CONCLUSION**

So which approach is “best”? In my opinion, it all depends on your personal comfort with algebraic manipulations. Some prefer to just take a derivative from the given form of . I avoid the more complicated quotient rule whenever I can, so the variation from line 4 would have been my approach.

The cool part is that it doesn’t matter what approach you use, so long as your algebraic manipulations are sound. **You don’t have to accept the form in which a problem is given; CHANGE IT to a form that works for you!**