# `xsin(y) = ycos(x)` Find `dy/dx` by implicit differentiation.

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### 1 Answer

Note:- 1) If y = sinx; then dy/dx = cosx

2) If y = cos(x) ; then dy/dx = -sinx

Now,

`x*sin(y) = y*cos(x)`

`or, x*cosy*(dy/dx) + sin(y) = -y*sin(x) + cos(x)*(dy/dx)`

`or, x*cos(y)*(dy/dx) - cos(x)*(dy/dx) = -y*sin(x) - sin(y)`

`or, (dy/dx)*[x*cos(y) - cos(x)] = -[y*sin(x) + sin(y)]`

`or, -(dy/dx)*[cos(x)-x*cos(y)] = -[y*sin(x)+sin(y)]`

`or, dy/dx = [y*sin(x) + sin(y)]/[cos(x) - x*cos(y)]`

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