# implicit differentiation - basic question from lang

I'm reading Lang's First Course in Calculus 5e. I am stuck on p. 104, where he introduces Implicit Differentiation. Here's the text:

"Find the derivative $dy/dx$ if $x^2 + y^2=7$, in terms of $x$ and $y$. We differentiate both sides of the equation using the chain rule, and the fact that $dx/dx=1$. We then obtain:

$2x\frac{dx}{dx} + 2y\frac{dy}{dx}=0$, that is $2x + 2y\frac{dy}{dx}=0$

So my question is totally basic: what does $\frac{dx}{dx}$ stand for? What's it doing in there in the first place? I thought $x^2$ could be differentiated directly to $2x$, without any intermediate steps, and I'm just not seeing what $\frac{dx}{dx}$ is doing in there. Thanks for any tips.

• I think he is just be super crystal clear with what step is occuring: he is regarding everything on the LHS as a function of x, and differentiating with the chain rule even though for that first term, its really not needed. Commented Sep 22, 2011 at 5:15
• You are right, at this stage it causes unnecessary confusion. But if we were moving around the circle $x^2+y^2=7$, and $x$ and $y$ were both functions of time $t$, we would have $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$. However, that's for later! Commented Sep 22, 2011 at 5:20
• I appreciate the nudges in the right direction, but I'm still head-scratching. How about this: how would you say $2x\frac{dx}{dx}$ in English? I think this is one of the things I'm really missing out on by studying on my own rather than in a class... Commented Sep 22, 2011 at 5:29

The $\frac{dx}{dx}$ is due to the chain rule. It means what other differentials mean: the rate of change of $x$ with respect to the rate of change of... well, $x$ in this case!
this ratio is equal to $1$
And yes, the derivative of $x^2$, with respect to $x$, is $2x$.
• Anytime! :) Commented Sep 22, 2011 at 13:14