Our teacher tried to explain to us how to find a slope at a given point of the function $y^2 = 2px$ by taking derivative from both sides of the equation, he did this: $$y^2= 2px $$ $$2yy' = 2p$$ $$ y' = \frac{p}{y}$$ I haven't quite understood the second part of the process, will be thankful for a logical and clear explanation :)

  • $\begingroup$ $y$ is a function in $x$, thus by chain rule $y^2=y^2(x)=2y(x)y'(x)$. $\endgroup$ – Frank Lu Jan 5 '17 at 6:36
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    $\begingroup$ The funny thing is that when you differentiate, for example $x^2$, what you are actually doing is the same: $$y=x^2 \\ [y]=[x^2] \\ y'=2x$$. $\endgroup$ – Billy Rubina Jan 5 '17 at 9:07
  • $\begingroup$ @Oppa Hilbert Style helped me the most, I've understood it a momet after I've sent the question ^_^ $\endgroup$ – Ozk Jan 5 '17 at 11:24

When we find derivative of some term. First we find the derivative of the power then the derivative of term.

So left hand side of second step,

$\frac{d}{dx}y^2 = 2y^{2-1}.\frac{d}{dx}y$

= $2y^{1}.\frac{dy}{dx}$

And you can write $\frac{dy}{dx} = y'$

So we have,

= $2y.y'$

In second step on the right hand side,


= 2p$\frac{d}{dx}x$

As 2 and p is constant and derivative of x with respect to x is 1.

= 2p

At last,

2y.y' = 2p.

$y' = \frac{p}{y}$

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    $\begingroup$ find this explanation very useful and coherent, tho you forgot the 2 in the 2px. $\endgroup$ – Ozk Jan 5 '17 at 11:28
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    $\begingroup$ @Ozk yes sorry for that. I edit it. $\endgroup$ – Kanwaljit Singh Jan 5 '17 at 12:14

It will become clear if you know the concept of implicit differentiation.

Thus we have that $$y^2=2px $$ $$\Rightarrow 2y\frac {dy}{dx} =2p\frac {dx}{dx} $$ $$\Rightarrow 2yy'= 2p $$ $$\Rightarrow y'=\frac {p}{y} $$ where $y^{(n)} $ is the nth derivative of $y$ (we generally use $y', y''$ etc. for lower derivatives but it is not advisable for higher ones). Hope it helps.


This is just the chain rule you presumably learned about earlier in calc, but the notation is more compact. Recall the chain rule says that $$ \frac{d}{dx}f(y(x)) = f'(y(x))y'(x) $$

The LHS of the second line is just the chain rule applied for $f(y) = y^2$. Since $f'(y) = 2y,$ we have $$ \frac{d}{dx}(y(x))^2 = 2y(x)y'(x). $$


$y$ is a function of $x$. $y^2$ is differentiated with respect to independent variable $x$ using Chain Rule. Instead of everytime writing like the following

$$y(x)^2= 2px, \,2y(x)y'(x) = 2p,\, y^{\prime}(x) = \frac{p}{y(x)}$$

your teacher has omitted repetitive parentheses $(x)$, assuming that you can appreciate that $p$ is a constant and $x,y$ are variables.

BTW $p$ is double the focal length of parabola which has a property that sub-normal ( projection on x-axis $=p)$ is always constant for the normal (in red).

enter image description here


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