if cos(4x-y)=x+y, then dy/dx=

Answer 1

To implicitly differentiate the equation, first we take the derivative of both sides of the equation with respect to x. We have

d/dx(cos(4x-y)) = d/dx(x+y).

On the LHS, since we have a function (4x-y) inside another function (cos), we use the chain rule and get

d/dx(cos(4x-y)) = -sin(4x-y) * (4-dy/dx) = -4sin(4x-y) + dy/dx(sin(4x-y)).

Apply the power rule to the RHS, we have

d/dx(x+y) = 1 + dy/dx.

Therefore, we have

-4sin(4x-y) + dy/dx (sin(4x-y)) = 1 + dy/dx.

Solving for dy/dx, we get

dy/dx (sin(4x-y)) - dy/dx = 1 + 4sin(4x-y)

dy/dx [sin(4x-y) - 1] = 1 + 4sin(4x-y)

dy/dx = [1 + 4sin(4x-y)]/[sin(4x-y) -1]