Respuesta :
I have a solution here for a slightly similar given which is ((sin(pix)+(cos(piy))^2 = 2. However, if you just analyze the steps and processes in the solution, you might just answer this problem on your own!
You need to remember here that pi is just a number
so sin(pix) differentiates to to pi*cos(pi*x)
(cos(pi*y))^2
the differentiate of this with respect to x goes to
2(cos(pi*y))* (-pi * sin(pi*y)) * dy/dx
2 differentiated with respect to x goes to 0
so you're left with
piCos(Pi(x)) -2Pi(dy/dx)Sin(Pi(y))Cos(Pi(y)) = 0
-2Pi(dy/dx)Sin(Pi(y))Cos(Pi(y)) = -piCos(Pi(x))
dy/dx = (-piCos(Pi(x))) / -2PiSin(Pi(y))Cos(Pi(y))
negatives and pi cancel so
dy/dx = (Cos(Pi(x))) / 2Sin(Pi(y))Cos(Pi(y))
I hope my guide has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
You need to remember here that pi is just a number
so sin(pix) differentiates to to pi*cos(pi*x)
(cos(pi*y))^2
the differentiate of this with respect to x goes to
2(cos(pi*y))* (-pi * sin(pi*y)) * dy/dx
2 differentiated with respect to x goes to 0
so you're left with
piCos(Pi(x)) -2Pi(dy/dx)Sin(Pi(y))Cos(Pi(y)) = 0
-2Pi(dy/dx)Sin(Pi(y))Cos(Pi(y)) = -piCos(Pi(x))
dy/dx = (-piCos(Pi(x))) / -2PiSin(Pi(y))Cos(Pi(y))
negatives and pi cancel so
dy/dx = (Cos(Pi(x))) / 2Sin(Pi(y))Cos(Pi(y))
I hope my guide has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
