implicit differentiation trig - peer check

[jinh0o]

Hello, I just tried to solve this equation and would like to see if I am headed towards the correct direction.


cos^2(x)+cos^2(y)=Cos(2x+2y)

=2(cosx)(-sinx)+2(cosy)(-siny)y'=-sin(2x+2y)(2+2y')

=2(cosx)(-sinx)+2(cosy)(-siny)y'=[-2sin(2x+2y)][-2sin(2x+2y)](y')

=2(cosy)(-siny)y'+2sin(2x+2y)y'=-2sin(2x+2y)-2(cosx)(-sinx)

=y'[2(cosx)(-sinx)+2sin(2x+2y)]=-2sin(2x+2y)-2(cosx)(-sinx)


y'=[-2sin(2x+2y)-2(cosx)(-sinx)]/[2(cosx)(-sinx)+2sin(2x+2y)]

Thank you!

 

[Deleted member 4993]

Hello, I just tried to solve this equation and would like to see if I am headed towards the correct direction.


cos^2(x)+cos^2(y)=Cos(2x+2y)

=2(cosx)(-sinx)+2(cosy)(-siny)y'=-sin(2x+2y)(2+2y')

=2(cosx)(-sinx)+2(cosy)(-siny)y'=[-2sin(2x+2y)][-2sin(2x+2y)](y')

=2(cosy)(-siny)y'+2sin(2x+2y)y'=-2sin(2x+2y)-2(cosx)(-sinx)

=y'[2(cosx)(-sinx)+2sin(2x+2y)]=-2sin(2x+2y)-2(cosx)(-sinx)


y'=[-2sin(2x+2y)-2(cosx)(-sinx)]/[2(cosx)(-sinx)+2sin(2x+2y)]

Thank you!

Your "question" is not very clear.

Are you trying "solve" for the value of x where this equation is valid?

Or

are you trying to derive the expression for $\displaystyle \frac{dy}{dx}$?

 

[stapel]

Hello, I just tried to solve this equation and would like to see if I am headed towards the correct direction.

Your subject line mentioned implicit differentiation. Are you attempting to prove an identity, solve an equation (but for which variable?), do an implicit differentiation (but with respect to what variable?), or something else?

cos^2(x)+cos^2(y)=Cos(2x+2y)

Is there any significance to the upper-case "C" on the right-hand side of the equation?

Thank you!

 

[Ishuda]

Hello, I just tried to solve this equation and would like to see if I am headed towards the correct direction.


cos^2(x)+cos^2(y)=Cos(2x+2y)

=2(cosx)(-sinx)+2(cosy)(-siny)y'=-sin(2x+2y)(2+2y')

=2(cosx)(-sinx)+2(cosy)(-siny)y'=[-2sin(2x+2y)][-2sin(2x+2y)](y')

=2(cosy)(-siny)y'+2sin(2x+2y)y'=-2sin(2x+2y)-2(cosx)(-sinx)

=y'[2(cosx)(-sinx)+2sin(2x+2y)]=-2sin(2x+2y)-2(cosx)(-sinx)


y'=[-2sin(2x+2y)-2(cosx)(-sinx)]/[2(cosx)(-sinx)+2sin(2x+2y)]

Thank you!

If you meant this as an implicit function problem, then you have made two mistakes. They appear to both be typo's. The bracket "[-2..." should be "-[2..." and the "x" should be "y". Otherwise everything else appears correct although you should simplify your solution and make a statement about the domain of validity, i.e.
$\displaystyle y^{'}(x)\, =\, -\dfrac{cos(y)\, sin(y)\, -\, sin(2x+2y)}{cos(x)\, sin(x)\, -\, sin(2x+2y)}$
There is an obvious 'simplification'
$\displaystyle y^{'}(x)\, =\, -\dfrac{sin(2y) - 2 sin(2x+2y)}{sin(2x) - 2sin(2x+2y)}$
and possibly others which I haven't looked into. When is the above equations valid?