Related Rates Calculus - Confused About What dx/dt, dy/dt and dx/dy mean

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Hello, I am confused as to how they got 5y in this problem when they multiplied dx/dy by dy/dt in the fourth line.

I am also confused as to what dx/dt and dx/dy and dy/dt mean.

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It would have been more obvious if that had inserted a line after line 3 which read:

$$\frac{dx}{dy}=y $$

Do you see why? (just differentiate line 3 w.r.t y).

They told you $$\frac{dy}{dt}=5$$ so line 5 is just putting the values in for each term.

If you look back into the history of math, there is a fascinating distinction of notation between Lagrange and Leibnitz. Lagrange wrote $$y=f(x)$$and $$y'=f'(x)$$ Leibnitz came up with the alternate notation of $$y=f(x)$$ so $$\frac{dy}{dx}=d\frac{f(y)}{dx}$$ Leibnitz' notation allows a lot more 'clever' use of fraction-like quantities, which usually work out - but can lead to problems if things aren't quite right (but you must be careful!).

See https://en.wikipedia.org/wiki/Leibniz%27s_notation

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$\frac{dy}{dt}$ and $\frac{dx}{dt}$ are just another way to denote derivatives of functions. $\frac{dy}{dt}$ is equivalent to $y'(t)$. Likewise, $\frac{dx}{dt}$ means the same thing as $x'(t)$. This notation is called Leibniz's notation because it was Leibniz who introduced it.

If you differentiate both sides of the equation $y^2-1=2x$ with respect to $t$ (since $x$ and $y$ are functions of $t$), you will get this:

$$2y\frac{dy}{dt}=2\frac{dx}{dt}\implies \frac{dx}{dt}=y\frac{dy}{dt}$$

$\frac{dy}{dt}$ is given. You also know that $x=12$. This means that at some point $t_0$, $x$ is equal to $12$. But since $y$ is a function of the same quantity, what would it be equal to at the same point $t_0$ when $x$ is $12$? Well, use the relationship between $y$ and $x$ that you're given—$y(t)=\sqrt{2x(t)+1}$: $$ y(t_0)=\sqrt{2x(t_0)+1}\\ y(t_0)=\sqrt{2\cdot 12+1}\\ y(t_0)=5\\y=5 $$

Just plug all the things that you now know into the expression that we came up with a few lines above:

$$\frac{dx}{dt}=5\cdot 5 = 25$$