Find (dy/dx) by implicit differentiation?

Question

5sinx x + 7cos y = sin x cos y

Sorry for the typo its 5sin x + 7cos y = sin x cos y

Thanks

Answer 1

Is that x next to the sin(x)? I'm going to assume it's a typo, because by convention, x is usually written in front of sin(x).

5sin(x) + 7cos(y) = sin(x)cos(y)

Differentiate, keeping in mind that when using the chain rule, the derivative of y is dy/dx.

5cos(x) + 7(-sin(y))(dy/dx) = cos(x)cos(y) + sin(x)(-sin(y))(dy/dx)

Move everything with a dy/dx to the left hand side; everything else goes to the right hand side.

7(-sin(y))(dy/dx) - sin(x)(-sin(y))(dy/dx) = -5cos(x) + cos(x)cos(y)

Simplify,

-7sin(y) (dy/dx) + sin(x)sin(y) (dy/dx) = -5cos(x) + cos(x)cos(y)

Factor (dy/dx).

(dy/dx) [ -7sin(y) + sin(x)sin(y) ] = -5cos(x) + cos(x)cos(y)

Divide both sides to isolate dy/dx.

dy/dx = [ -5cos(x) + cos(x)cos(y) ] / [ -7sin(y) + sin(x)sin(y) ]

Factor the top and bottom.

dy/dx = [cos(x) (-5 + cos(y))] / [ sin(y) [-7 + sin(x)] ]

dy/dx = [cos(x)/sin(y)] [ (-5 + cos(y)) / (-7 + sin(x)) ]

Answer 2

Diff. both sides with respect to x,
5 cos x - 7siny.dy/dx = cosx.cosy - siny.sinx.dy/dx
simplifying,
dy/dx = (5cosx - cosx.cosy)/(7siny-siny.sinx)