# Use implicit differentiation to determine dy/dx given that sin x*cos y = tan y

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The derivative `dy/dx` has to be determined given that sin x*cos y = tan y.

Take the derivative of both the sides with respect to x

`(d(sin x*cos y))/dx = (d(tan y))/dx`

=> `cos x - sin y*(dy/dx) = sec^2 y*(dy/dx)`

=> `(dy/dx)(sec^2y + sin y) = cos x`

=> `(dy/dx) = (cos x)/(sec^2y + sin y)`

**The required derivative is `(dy/dx) = (cos x)/(sec^2y + sin y)` **