How do you find $dy/dx$ by implicit differentiation of $(sinpix+cospiy)^2=2$ ?

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Answer 1 · 2017-01-27T22:07:50

Please see the explanation.

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

Differentiate:

$(d((sin(pix)+cos(piy))^2))/dx = (d(2))/dx$

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

The chain rule is:

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

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

$(dh)/dx = picos(pix) -pisin(piy)dy/dx$

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

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

$(d((sin(pix)+cos(piy))^2))/dx = 2h(x,y)(picos(pix) -pisin(piy)dy/dx)$

$(d((sin(pix)+cos(piy))^2))/dx = 2(sin(pix)+cos(piy))(picos(pix) -pisin(piy)dy/dx)$

The right side is just derivative of a constant:

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

Put the terms back into the equation:

$2(sin(pix)+cos(piy))(picos(pix) -pisin(piy)dy/dx) = 0$

Solve for $dy/dx$

$-cos(pix) + sin(piy)dy/dx = 0$

$sin(piy)dy/dx = cos(pix)$

$dy/dx = cos(pix)/sin(piy)$

How do you find dy/dxdy/dx by implicit differentiation of (sinpix+cospiy)^2=2(sinpix+cospiy)^2=2?