implicit differentiation (y^2)/36
By Sarah Smith
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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?
$\endgroup$ 14 Answers
$\begingroup$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 $$\frac{d}{dx}\left(\frac{x^2}{25}\right)=\frac{1}{25}*\frac{d}{dx}\left(x^2\right)=\frac{1}{25}*2x\frac{d}{dx}\left(x\right)=\frac{2x}{25}*1=\frac{2x}{25}.$$
$\endgroup$ 6 $\begingroup$That is essentially what wolfram alpha would show to its user. If you were to calculate this on mathematica 9.0. You would get:
which basically is what you expected. Hope this helps.
$\endgroup$ 2 $\begingroup$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
$$ \begin{align*} y' = \frac{dy}{dx} & = \frac{36x}{25y} \end{align*} $$
$\endgroup$ 1 $\begingroup$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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