# Differentiation/ implicit differentiation question

Confused about how these two equate to each other:

$$\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{dy}{dx}\frac{d}{dy}\left(\frac{dy}{dx}\right)$$

Would anyone kindly explain how to get from the left side to the right? (The left side being the second derivative with respect to x, $$\frac{d^2y}{dx^2}$$)

I'm unsure what exactly differentiating with respect to y does for the equation.

Many thanks

• Isn’t this just an application of the chain rule?
– amd
Commented Dec 20, 2019 at 20:17

The way to understand it is to recognize that most treatments of higher-order differentials are fundamentally misleading (I would say wrong, but it makes people snippy when I do that). The second derivative is not $$\frac{d^2y}{dx^2}$$!

If you actually take the derivative of the derivative, you would have to recognize that the first derivative is actually a fraction, and therefore the appropriate tool to use is the quotient rule. It's actually two parts - taking a differential and then dividing by $$dx$$.

So, just the differential gives:

$$d\left(\frac{dy}{dx}\right) = \frac{dx\,d^2y - dy\,d^2x}{dx^2}$$ Then divide by $$dx$$: $$\frac{d\left(\frac{dy}{dx}\right)}{dx} = \frac{dx\,d^2y - dy\,d^2x}{dx^3}$$ Then I like to simplify it like this: $$\frac{d\left(\frac{dy}{dx}\right)}{dx} = \frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$$

That is the left-hand side of your question. On the right-hand side, you have the differential (like the first operation), but then divide by $$dy$$, but then you multiply by $$\frac{dy}{dx}$$, which, when you combine all those operations together, give you the same result as just having divided by $$dx$$ in the first place.

So, $$\frac{dy}{dx} \cdot \frac{d\left(\frac{dy}{dx}\right)}{dy} = \frac{dy}{dx}\cdot \frac{\frac{dx\,d^2y - dy\,d^2x}{dx^2}}{dy} = \frac{\frac{dx\,d^2y - dy\,d^2x}{dx^2}}{dx} = \frac{dx\,d^2y - dy\,d^2x}{dx^3} = \frac{d\left(\frac{dy}{dx}\right)}{dx} = \frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$$

Which is the same value as before.

Another, perhaps simpler, way of doing this, is just separating the differential from the derivative in the first place. So, we would rewrite your original equation as:

$$\frac{d\left(\frac{dy}{dx}\right)}{dx} = \frac{dy}{dx} \cdot \frac{d\left(\frac{dy}{dx}\right)}{dy}$$

Which is much more obviously true, but it requires separating thinking about the derivative and the differential.

For more information on this, see the papers "Extending the Algebraic Manipulability of Differentials," "Simplifying and Refactoring Introductory Calculus," or Appendix B of Calculus from the Ground Up.

• Many thanks for this, it's an interesting explanation that one spent some time trying to understand Commented Dec 24, 2019 at 13:25