How do you find dy/dxdydx by implicit differentiation of (sinpix+cospiy)^2=2(sinπx+cosπy)2=2?

1 Answer
Jan 27, 2017

Please see the explanation.

Explanation:

Given: (sin(pix)+cos(piy))^2 = 2(sin(πx)+cos(πy))2=2

Differentiate:

(d((sin(pix)+cos(piy))^2))/dx = (d(2))/dxd((sin(πx)+cos(πy))2)dx=d(2)dx

Use the chain rule on (d((sin(pix)+cos(piy))^2))/dxd((sin(πx)+cos(πy))2)dx

The chain rule is:

(d(g(h(x,y))))/dx = (dg)/(dh)(dh)/dxd(g(h(x,y)))dx=dgdhdhdx

let h(x,y) = sin(pix)+cos(piy)h(x,y)=sin(πx)+cos(πy), then:

(dh)/dx = picos(pix) -pisin(piy)dy/dxdhdx=πcos(πx)πsin(πy)dydx

g = (h(x,y))^2g=(h(x,y))2

(dg)/(dh) = 2h(x,y)dgdh=2h(x,y)

(d((sin(pix)+cos(piy))^2))/dx = 2h(x,y)(picos(pix) -pisin(piy)dy/dx)d((sin(πx)+cos(πy))2)dx=2h(x,y)(πcos(πx)πsin(πy)dydx)

(d((sin(pix)+cos(piy))^2))/dx = 2(sin(pix)+cos(piy))(picos(pix) -pisin(piy)dy/dx)d((sin(πx)+cos(πy))2)dx=2(sin(πx)+cos(πy))(πcos(πx)πsin(πy)dydx)

The right side is just derivative of a constant:

(d(2))/dx = 0d(2)dx=0

Put the terms back into the equation:

2(sin(pix)+cos(piy))(picos(pix) -pisin(piy)dy/dx) = 02(sin(πx)+cos(πy))(πcos(πx)πsin(πy)dydx)=0

Solve for dy/dxdydx

-cos(pix) + sin(piy)dy/dx = 0cos(πx)+sin(πy)dydx=0

sin(piy)dy/dx = cos(pix)sin(πy)dydx=cos(πx)

dy/dx = cos(pix)/sin(piy)dydx=cos(πx)sin(πy)