2
$\begingroup$

Why, after differentiating $y$ on one side of the equation, is $dy/dx$ added?
As clarification an example I will provide an example:

Implicitly differentiate $y^2 = x$.

You get $$2y\frac {dy}{dx} = 1$$ as one of the first steps in differentiation. Why is the $dy/dx$ added after $y^2$ is differentiated?

Thanks!

Improve this question
$\endgroup$
  • 1
    $\begingroup$ because of the chain rule, $y$ is a function of $x$ and so the derivative of $y$ is $1$ multiplied by its derivative, which is $y'$ $\endgroup$ – imranfat Apr 9 '14 at 19:49
  • 1
    $\begingroup$ @Ethan $y^2=x$ is abuse of notation for $\left(y(x)\right)^2=x$. Just apply the chain rule. $\endgroup$ – Git Gud Apr 9 '14 at 19:50
1
$\begingroup$

$$y^2=(y(x))^2=x\\ \dfrac{d(y^2)}{dx}=1\\ \implies \dfrac{d(y^2)}{dx}=\dfrac{d(y^2)}{dy}\dfrac{dy}{dx}\text{ (chain rule )}=1\\ \implies 2y\dfrac{dy}{dx}=1$$

Improve this answer
$\endgroup$
  • $\begingroup$ When you get $\frac{d(y^2)}{dy} \frac{dy}{dx}$, wouldn't the $dy$ cancel leaving $\frac{d(y^2)}{dx} = 2y$ ? $\endgroup$ – OpieDopee Apr 9 '14 at 20:21
  • $\begingroup$ @Ethan $$\dfrac{d(y^2)}{dy}\dfrac{dy}{dx}=2y\boxed{\dfrac{dy}{dx}}$$Of course, you can cancel out the $dy$s, but what do you get? You get $\dfrac{d(y^2)}{dx}$, which is what you began with. $\endgroup$ – user122283 Apr 9 '14 at 20:25
  • $\begingroup$ So because there are no $x$s in $\frac{d(y^2)}{dx}$, you can differentiate with respect to $y$ giving you $2y$ ? $\endgroup$ – OpieDopee Apr 9 '14 at 20:29
  • $\begingroup$ @Ethan Let me give an example. Let $y=x+1$. Then, $y^2=x^2+1+2x$. Then, $$\dfrac{d(y^2)}{dx}=\dfrac{d(x^2+1+2x)}{dx}=2x+2\\ \dfrac{d(y^2)}{dx}=\dfrac{d(y^2)}{dy}\dfrac{dy}{dx}=2y\dfrac{dy}{dx}=2(x+1)(1)=2x+2$$See that both are equal. $\endgroup$ – user122283 Apr 9 '14 at 20:34
  • $\begingroup$ In the second half of the example when you get $2y \frac{dy}{dx}$ how did $\frac{dy}{dx}$ simplifying to $1$? $\endgroup$ – OpieDopee Apr 9 '14 at 20:42
1
$\begingroup$

The idea behind implicit differentiation is that while $y$ isn't (necessarily) a function of $x$ we treat it as it it was.

So say that $y = f(x)$. Then you want to find $f'(x)$ in the following $$f(x)^2 = x.$$ So you take the derivative on both sides and use the chain rule to find the derivative of $f(x)^2$: $$\begin{align} \frac{d}{dx} f(x)^2 &= \frac{d}{dx} x & \Rightarrow \\ 2f(x)f'(x) &= 1 \end{align} $$ Now then we just replace $f(x)$ by $y$ and get $$ 2y\frac{dy}{dx} = 1. $$ So to answer you question about where the $\frac{dy}{dx}$ comes from, you can think of it as the derivative of the inner function $y$.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.