Question

Use implicit differentiation to find an equation of the tangent line to the graph at the given point.
y^2 +ln(xy) = 2, (e, 1)

Answer 1

start with
\[2yy'+\frac{1}{xy}\left(y+xy'\right)=0\]

Answer 2

Makes sense so far.

Answer 3

first part is clear i hope, second is chain rule plus product rule

now you have a choice: you can solve for \(y'\) using algebra, then replace \(x\) by \(e\) and \(y\) by \(1\) to get the slope, or you can make the replacement at this step and then solve for \(y'\)

Answer 4

Could you show me the first way?

Answer 5

i bunch of algebra, easier to do it with pencil and paper

Answer 6

Alright, that is fine. I should be able to get it from here. Thank for the help!

Answer 7

i can try
distribute first
\[2yy'+\frac{1}{x}+\frac{y'}{y}=0\] then
\[2yy'+\frac{y'}{y}=-\frac{1}{x}\]
\[y'(2y+\frac{1}{y})=-\frac{1}{x}\]
then divide
hope my algebra is good

Answer 8

So then the tangent line would be (1-y) = (-1/x)/(2y+(1/y))(e-x) ?

Answer 9

there should be no \(x\) or (y\) in your answer \(x=e,y=1\)

Answer 10

oh is see, the equation for the line, sorry

Answer 11

\[y-1=m(x-e)\] for whatever ugly expression you got for \(m\)

Answer 12

but \(m\) is the slope, it should be a number

Answer 13

lets start at this line here
\[y'(2y+\frac{1}{y})=-\frac{1}{x}\] now \(x=e,y=1\) so you get
\[3y'=-\frac{1}{e}\] and therefore \(y'=-\frac{1}{3e}\) is your slope, assuming my algebra was right
i is a number, there is no variable in the slope

Answer 14

Alright, I'll go with it, thank you so much for the help!