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?
Best Answer
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 $$\frac{d}{dx}\left(\frac{x^2}{25}\right)=\frac{1}{25}*\frac{d}{dx}\left(x^2\right)=\frac{1}{25}*2x\frac{d}{dx}\left(x\right)=\frac{2x}{25}*1=\frac{2x}{25}.$$