I'm currently enrolled in Calculus 1, and everything has been pretty smooth up until these last two sections involving the chain rule and implicit derivation. After watching multiple YouTube videos, and reading different professors' explanations, all of these resources either explain it in a different way, or use different notations/process to get to the answer. I'm really lost, and would like help on this problem I'm working on and if possible, an explanation of what is going on with implicit derivation/how to solve these problems.
$$ y = \frac{x+5}{y-5}$$ $$ \frac{dy}{dx} = ? $$
So far I've tried two approaches:
1st approach:
$(y-5)y = x+5$
Taking deriv of both sides
$2ydy - 5 = xdx$
$2ydy=xdx+5$
$dy = \frac{xdx + 5}{2y}$
$\frac{dy}{dx} = \frac{x+5}{2y}$
2nd approach:
$dy = \frac{\frac{d}{dx}(x+5)(y-5) - (x+5)(\frac{d}{dy}(y-5))}{(y-5)^2}$
$dy = \frac{xdx(y-5) - (x+5)(ydy)}{(y-5)^2}$
Then I tried multiplying a (y-5) out from the numerator, resulting in the left side $dy(y-5)$, and got stuck.
If someone could please explain to me where my lapses in understanding are, or help me in any way that would be amazing. Thank you.
This is one you can solve two ways to check the work.
Implicit:
$y^2-5y = x+5$
$2y\frac{dy}{dx}-5\frac{dy}{dx}=1$
$\frac{dy}{dx}=\frac{1}{2y-5}$
And we are finished.
But double check:
$y^2-5y = x+5$
$(y-\frac{5}{2})^2 = \frac{25}{4}+x+5$
$y=\frac{5}{2} \pm \sqrt{\frac{45}{4}+x}$
$\frac{dy}{dx}=\frac{1}{2} (\frac{45}{4}+x)^{-\frac{1}{2}} = \frac{1}{2\sqrt{\frac{45}{4}+x}}$
Return to the implicit and plug in the $y$ we just solved for:
$\frac{dy}{dx}=\frac{1}{2y-5}$
$\frac{dy}{dx}=\frac{1}{2\Big( \frac{5}{2} \pm \sqrt{\frac{45}{4}+x} \Big)-5} = \frac{1}{2\sqrt{\frac{45}{4}+x}}$ as desired.
Rather than professors notes, or YouTube videos, I might suggest reading the sections of one or two textbooks on the topic, it will likely clear things up for you.