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Find $$\frac{dy}{dx}$$ where y and x satisfy the implicit equation: $$ \displaystyle \quad\quad \frac{x^2}{25} + \frac{y^2}{36} = 1. $$

Now, I'm having trouble with finding the derivative for $$\frac{y^2}{36}$$

I know that when doing implicit differentiation I need to write y as a dependent of x, thus, I'm differentiating: $$\frac{(y(x))^2}{36}$$ so if you differentiate it by either quotient rule or chain rule you get: $$ \frac {1}{18} y(x) * y'(x) $$

but how come, wolfram alpha yields almost an identical answer yet without (x) in the second term?

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    $\begingroup$ $y$ is a function of $x$, hence it is correct to write $y(x)$. However, this notations can become quite cumbersome when differentiating, especially with more complicated expressions, so the convention is to write $y$ instead of $y(x)$. $y'$ is therefore understood to mean $y'(x)$. $\endgroup$ – Reveillark Feb 22 '15 at 3:35
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That is essentially what wolfram alpha would show to its user. If you were to calculate this on mathematica 9.0. You would get:

Implicit Differentiation

which basically is what you expected. Hope this helps.

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  • $\begingroup$ Okay, your answer is the one that addresses my problem specifically. So in short, I got the correct answer, it's just that wolfram alpha doesn't show it properly. $\endgroup$ – dramadeur Feb 22 '15 at 4:15
  • $\begingroup$ Yes, you got the point. $\endgroup$ – Terrence Liao Feb 22 '15 at 5:39
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Taking the derivative of $\displaystyle\frac{y^2}{36}$, we have $$\frac{d}{dx}\left(\frac{y^2}{36}\right)=\frac{d}{dx}\left(\frac{1}{36}*y^2\right)=\frac{1}{36}*\frac{d}{dx}\left(y^2\right)=\frac{1}{36}*2y\frac{d}{dx}\left(y\right)=\frac{2y}{36}*\frac{dy}{dx}=\frac{y}{18}*\frac{dy}{dx}.$$

In the step where we take the derivative of $y$, since $y$ is a function in terms of $x$, it's derivative $\displaystyle\frac{dy}{dx}$ represents the change in $y$ with respect to $x$.

To answer your question, as to why the second term in your WolframpAlpha query is simply $\displaystyle\frac{2x}{25}$, notice what $\displaystyle\frac{d}{dx}(x)=\frac{dx}{dx}$ represents: $\textbf{the change in x with respect to x}$, which is $1$.

So, taking the derivative of $\displaystyle\frac{x^2}{25}$, we have $$\frac{d}{dx}\left(\frac{x^2}{25}\right)=\frac{1}{25}*\frac{d}{dx}\left(x^2\right)=\frac{1}{25}*2x\frac{d}{dx}\left(x\right)=\frac{2x}{25}*1=\frac{2x}{25}.$$

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  • $\begingroup$ But that's the point, it's not just "y^2/36", it's: (y(x))^2)/36, and the derivative is: 1/18* y(x)∗y′(x). I think you missed the fact that we're taking derivative in respect to x. $\endgroup$ – dramadeur Feb 22 '15 at 4:09
  • $\begingroup$ y should be written as a dependent of x. So you replace y as y(x). So what is this $$\frac{dy}{dx}$$ is it the same as y'(x)? $\endgroup$ – dramadeur Feb 22 '15 at 6:46
  • $\begingroup$ @dramadeur yeah it's the same, i'm just answering one of your questions: "but how come, wolfram alpha yields almost an identical answer yet without (x) in the second term?", as I thought you knew how to solve the problem already. $\endgroup$ – user144809 Feb 22 '15 at 6:47
  • $\begingroup$ I wasn't referring to 2x/25 I've got no problem with this. I was referring to wolfram giving me the answer: 1/18 y * y'(x), whereas it should be 1/18 y(x) * y'(x). $\endgroup$ – dramadeur Feb 22 '15 at 6:48
  • $\begingroup$ @dramadeur I also addressed that, i'm just using a different notation, but it's exactly what wolfram alpha has. (I can post an equivalent notation if it helps). $\endgroup$ – user144809 Feb 22 '15 at 6:49
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You basically need to apply the differentiation operator $\frac{d}{dx}$ on both sides of the equation:

$$ \begin{align*} \frac{d}{dx} \{ \frac{x^2}{25} + \frac{y^2}{36} \} & = \frac{d}{dx} \{ 1 \} \\ \frac{2x}{25} + \frac{2y y'}{36} & = 0 \\ \end{align*} $$

Solving for $y'$ gives us

$$ \begin{align*} y' = \frac{dy}{dx} & = \frac{36x}{25y} \end{align*} $$

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  • $\begingroup$ Uhm... I have no problem with the rest of the problem. I have trouble with its specific part, y^2/36. Now, you said the derivative of it is (2y*y')/36 ... I guess that's true once you remove (x)'s from y's. But before that, it's what I wrote, right? 1/18*y(x)∗y′(x) $\endgroup$ – dramadeur Feb 22 '15 at 4:14
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Here are the steps $$ \frac{x^2}{25} + \frac{y^2}{36} = 1$$ $$ \frac{d}{dx}\left[\frac{x^2}{25} + \frac{y^2}{36}\right] = \frac{d}{dx}[1]$$ $$ \frac{d}{dx}\left[\frac{x^2}{25}\right] + \frac{d}{dx}\left[\frac{y^2}{36}\right] = 0$$ $$ \frac{1}{25}\frac{d}{dx}\left[x^2\right] + \frac{1}{36}\frac{d}{dx}\left[y^2\right] = 0$$ $$ \frac{2x}{25} + \frac{2y}{36}\frac{d}{dx}\left[y\right] = 0$$ $$ \frac{y}{18}\frac{d}{dx}\left[y\right] = -\frac{2x}{25}$$ $$ y\frac{d}{dx}\left[y\right] = -\frac{36x}{25}$$ $$ \frac{d}{dx}\left[y\right] = -\frac{36x}{25y}$$

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  • $\begingroup$ You're not supposed to treat y^2/36 like this. The derivative of y^2/36 in respect to x is: 1/18 * y(x) * y'(x) $\endgroup$ – dramadeur Feb 22 '15 at 6:41
  • $\begingroup$ @dramadeur, that is exactly what I posted...I favor Leibniz's notation. $\endgroup$ – k170 Feb 22 '15 at 11:49
  • $\begingroup$ @dramadeur, you don't have to explicitly write $y(x)$ as it's quite obvious that $y$ is dependent on $x$ in these type of problems. $\endgroup$ – k170 Feb 22 '15 at 12:00

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