Arctan(y)=3x+y how would I go about finding y'?

[IllmicIll]

Homework Statement

Arctan(y)=3x+y how would I go about finding y'?

Homework Equations

The Attempt at a Solution

I tried to start out with...

arctan(y)= tan(x)
tan(x)=3x+y
tan(x)-3x=y

y'=1/(1+x^2) -3

is this correct?
thanks in advance.

 

[rootX]

arctan(y)= tan(x)

How you reached here from
Arctan(y)=3x+y ?

What's
d/dx (Arctan(y)) ?
Use the chain rule.

 

[Dick]

How can arctan(y)=tan(x) when your equation says arctan(y)=3x+y?? That's not right. Don't try to solve for y. You can't. Differentiate it implicitly.

 

[IllmicIll]

d/dx arctan y

y'/1+y^2*y' ?

 

[Dick]

d/dx arctan y

y'/1+y^2*y' ?

Close. But why are there two y' in there? And use parentheses to avoid confusion. There's a difference between 1/1+y^2 and 1/(1+y^2).

 

[IllmicIll]

y'/(1+y^2) *y'
I thought the 2nd y' has to be there bc for example
y=arccot(x^2)
x^2=cot y
2x=-csc^2(y) *y'

is that not the case since I already have a y'?

 

[IllmicIll]

oops nvm...