# Write an equation for the line tangent to the curve xsin(2y)=ycos(2x) at the point (pi/4,pi/2)

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### 1 Answer

The slope of a tangent to a curve f(x) at the point x = c is the value of f'(c).

It is given that `x*sin(2y)=y*cos(2x)` . Use implicit differentiation to find `dy/dx`

`sin(2y) + x*cos(2y)*2*(dy/dx) = (dy/dx)*cos(2x) + y*sin 2x*2`

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

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

At the point `(pi/4, pi/2)`

`dy/dx = (2*pi/2*1 - 0)/(pi/2*-1 - 0)`

`dy/dx = (pi)/(-pi/2)`

`dy/dx = -2`

The equation of the tangent is `(y - pi/2)/(x -pi/4) = 2`

=> `y - pi/2 = 2x - pi/2`

=> `y = 2x`

**The tangent to `x*sin(2y)=y*cos(2x)` at `( pi/4, pi/2)` is **`y = 2x `