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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 :)

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  • $\begingroup$ $y$ 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
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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,

$\frac{d}{dx}(2px)$

= 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
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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.

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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 $$ \frac{d}{dx}(y(x))^2 = 2y(x)y'(x). $$

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$y$ 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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