# Use Implicit Divverentiation to find an equation to the tangent line to the curve at the given point (pi,pi) sin(x+y)=2x-2y so far I have cos(x+y) * 1+y' = 2-2y' how do I get y' to one side? for...

Use Implicit Divverentiation to find an equation to the tangent line to the curve at the given point (pi,pi)

sin(x+y)=2x-2y

so far I have cos(x+y) * 1+y' = 2-2y'

how do I get y' to one side? for the left side isnt y' part of the whole equation so i have to keep that to the left side and bring -2y' over to the left also leaving 2 on the right???????

Step by step please!!!

### 3 Answers | Add Yours

Your first step is partially correct, you are missing the brackets around 1+y':

`sin(x+y)=2x-2y`

`cos(x+y)(1+y')=2-2y'`

Now we can expand and rearrange:

`cos(x+y)+y'cos(x+y)=2-2y'`

`y'cos(x+y)+2y'=2-cos(x+y)`

`y'(cos(x+y)+2)=2-cos(x+y)`

`y'=(2-cos(x+y))/(cos(x+y)+2)`

**Sources:**

`sin(x+y)=2x-2y`

`d/dx sin(x+y)=d/dx ( 2x-2y)`

`cos(x+y) xx d/dx (x+y)=2-2y'`

`cos(x+y)(1+y')=2-2y'`

`cos(x+y)+y'cos(x+y)=2-2y'`

`y'(cos(x+y)+2)=2-cos(x+y)`

`y'=(2-cos(x+y))/(cos(x+y)+2)`

Settings `x=pi` ; `y=pi`

`y'=1/3`

So: `y-pi=1/3(x-pi)`

`3y-3pi-x+pi=0`

`3y-x-2pi=0` tangent equation in point `P(pi;pi)`

Red line implicit projetion of tangent by the space.

Your first step is partially correct, you are missing the brackets around 1+y':

Now we can expand and rearrange:

What about finding the tangent line now?