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I was given the following problem:

Find $\frac{dy}{dx}$ using the implicit equation $x^2 + y^2 = 1$

What I'm more interested in is the explicit equation, $y = \sqrt{1 - x^2}$ (I'm allowed to assume only the positive part of the equation, I know it's actually $y = \pm\sqrt{1 - x^2}$). Lets say we tried to differentiate $y$ directly:

$$ \begin{align} \frac{dy}{dx} & = \frac{d}{dx}\sqrt{1 - x^2}\\ & = \frac{d}{dx}(1 - x^2)^{1/2}\\ & = 1/2(1 - x^2)^{-1/2}\\ & = \frac{1}{2\sqrt{1 - x^2}} \end{align} $$

Now this is clearly wrong. What we actually found was $\frac{dy}{d(1 - x^2)}$. So we have to use the chain rule to get $\frac{dy}{dx} = \frac{-x}{\sqrt{1 - x^2}}$.

Now lets take a look at $\frac{d}{dx}(1 - x^2)$:

$$ \frac{d}{dx}(1 - x^2) = -2x $$

Because $\frac{d}{dx}1 = 0$, and $\frac{d}{dx}(-x^2) = -2x$. But we can still technically consider $1 - x^2$ to be the composition of two functions $f(x) = 1 + x$ and $g(x) = x^2$. If we use the chain rule to find the derivative of $f \circ g$, we still see that it is equal to $-2x$.

So what is it about expressions like $\sqrt{1 - x^2}$ that don't allow us to differentiate them directly? Like what I said above about how we technically only found $\frac{dy}{d(1 - x^2)}$, what's to say that the way I differentiated $\frac{d}{dx}(1 - x^2)$ isn't actually the right way to do it because it is just analogous to $\frac{d}{d(1 + x)}$? What is the fundamental difference between two expressions that determines whether or not they can be differentiated directly, or if they need to be differentiated using the chain rule?

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  • $\begingroup$ You can do the second one "directly" because the derivative operator is linear, and you just have a sum (well, a difference). You can't do that with the first one because you have a non-linearity. $\endgroup$ – symplectomorphic Jul 18 '14 at 14:08
  • $\begingroup$ Linearity, like $f(x + y) = f(x) + f(y)$ & $f(kx) = kf(x)$, or some other property (or is that the only definition of linearity)? $\endgroup$ – user3002473 Jul 18 '14 at 14:10
  • $\begingroup$ The elementary formula $D(f+g)=Df+Dg$ is actually a special case of the chain rule, in which you introduce $S(u,v)=u+v$ and differentiate $S(f(\cdot),g(\cdot))$. Since $S$ is linear, it seems you don't see the chain rule, but it is there. $\endgroup$ – Siminore Jul 18 '14 at 14:12
  • $\begingroup$ What functions are you allowed to assume the derivative as 'known' for? $\endgroup$ – Winther Jul 18 '14 at 14:17
  • $\begingroup$ @Siminore: I disagree in the following sense: you can prove the linearity of the derivative without using the chain rule. of course you can then go back and see the linearity as a special case, after proving the chain rule. $\endgroup$ – symplectomorphic Jul 18 '14 at 14:17
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You can differentiate the second expression "directly" because of the power rule for monomials and because the derivative operator is linear: the derivative of a sum is the sum of the derivatives, and the derivative of a constant times a function is just the constant times the derivative of the function.

But you need the chain rule for the first expression because it cannot be written as a sum or difference of monomials.

The distinction isn't sharp, however. Consider $$f(x)=(2x+1)^{2014}$$

Technically you can rewrite this function as a sum, using the binomial theorem. So in theory you have no need for the chain rule. But in practice that would be terrible; the chain rule helps you get the derivative in factored form.

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  • $\begingroup$ Ah I see now. That makes more sense. $\endgroup$ – user3002473 Jul 18 '14 at 14:19
  • $\begingroup$ Oh! Okay, that makes total sense! Lol before I just thought it made sense, but I see it completely now! So that explains why you can't find things like $\frac{d}{dx}\sin{10x}$ without the chain rule, but why $\frac{d}{dx}(1 + x^2)$ can! Okay, that completely clears up my confusion. Thank you! $\endgroup$ – user3002473 Jul 18 '14 at 14:22
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    $\begingroup$ @user: technically you could find the derivative of $\sin10x$ using the definition of the derivative, instead of the chain rule. That's partly what Ross is getting at. But the chain rule does make life a little easier there. $\endgroup$ – symplectomorphic Jul 18 '14 at 14:33
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You start with a small list of derivatives that you derive from definitions and compound those with the various rules to cover (hopefully) all the functions you want to take the derivative of. When you try to take the derivative of $\sqrt {1-x^2}$, that isn't one you know already, so you have to break it down. If you do it often enough you might know that the derivative is $\frac{-x}{\sqrt{1-x^2}}$. You could claim this as a theorem and use it "directly". When you take the derivative of $1-x^2$ you really don't do it directly-the only basic form you proved from the definition was probably $\frac d{dx}x^n=nx^{n-1}$. It is just simple enough to use the sum rule to say $\frac d{dx}(1-x^2)=\frac d{dx}1+\frac d{dx}(-x^2)$ then the multiplication by constants to take this to $-2x$

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