implicit differentiation?

[Ric]

<Solved> implicit differentiation?

(x+y)^2=y-8 find the equation of the tangent line at (2,1).

I couldn't get y' alone =/

 

[Deleted member 4993]

(x+y)^2=y-8 find the equation of the tangent line at (2,1).

I couldn't get y' alone =/

2*(x+y)(1+y') = y'

y' = 2*(x+y)/[1-2*(x+y)]

at (2,1)

y' = 2*(2+1)/[1-2*(2+1)] = - 6/5

Now continue.....

 

[Ric]

2*(x+y)(1+y') = y'

***

y' = 2*(x+y)/[1-2*(x+y)]

How did you go from the first step to the 2nd?


edit: Is there a formula or theorem or...?

 

[pappus]

How did you go from the first step to the 2nd?


edit: Is there a formula or theorem or...?

Subhotosh Khan used the distributive law:

\(\displaystyle \begin{array}{r}2(x+y)(1+y') = y' \\ \implies~2(x+y)+2(x+y) \cdot y'=y' \\ \implies~ 2(x+y)=y'-2(x+y) \cdot y' \\ \implies~ 2(x+y)=y'(1-2(x+y)) \end{array}\)

Now divide by the factor of y'.

 

[Ric]

Thank you so much! I've never been taught that law before.

 

[Deleted member 4993]

Thank you so much! I've never been taught that law before.

Are you saying you haven't been taught the following:

a*(b+c) = a*b + a*c


????

And you are in Calculus????