Implicit Differentiation of xy-x-4+0 to find tangent at x=2

[stephanie953]

The problem asks find the equation(s) of the tangent line(s) to the graph of the indicated equations at the point(s) with the given value of x.

xy-x-4=0; x=2

First I solved to find what y was when x was 2

2y-2-4=0
2y=6
y=3

Then to find the slope I need to differentiate the equation implicitly:
I think this is where I went wrong...

xyy' -1=0
(xy)y'=1
y'=1/xy

Can someone help me figure out what I did wrong? Thank you.

 

[royhaas]

Re: Implicit Differentiation xy-x-4+0; x=2

The derivative equation is xy'+y-1=0.

 

[stephanie953]

Re: Implicit Differentiation xy-x-4+0; x=2

The derivative equation is xy'+y-1=0.

Thanks for the reply. So you split the xy into xy' + y?

 

[Deleted member 4993]

Re: Implicit Differentiation xy-x-4+0; x=2

Thanks for the reply. So you split the xy into xy' + y?

'xy' - is two functions multiplied with each other - it is like f(x) * y(x) ...... where f(x) = x and f'(x) = 1.

From there we apply "product rule of differentiation" - which is

\(\displaystyle \frac{d}{dx}[f(x)\cdot y(x)] \, = \, f'(x)\cdot y(x) \, + \, f(x)\cdot y'(x) \, = \, 1\cdot y(x) \,+ \, x\cdot y'(x) \, = \, y + \, x\cdot y'\)