# Use implicit differentiation to find an equation of the tangent line to the curve at the given point. ysin(16x) = xcos(2y)  point=(pi/2, pi/4)

For y = f(x), the slope of the tangent at any point (x, y) is the value of `dy/dx` at that point.

The curve is defined in the problem by `y*sin(16x) = x*cos(2y)`

Using implicit differentiation

`y*cos(16x)*16 + sin(16x)(dy/dx) = x*(-sin 2y)*2*(dy/dx) + cos 2y`

=> `(dy/dx)(sin 16x + 2x*sin...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

For y = f(x), the slope of the tangent at any point (x, y) is the value of `dy/dx` at that point.

The curve is defined in the problem by `y*sin(16x) = x*cos(2y)`

Using implicit differentiation

`y*cos(16x)*16 + sin(16x)(dy/dx) = x*(-sin 2y)*2*(dy/dx) + cos 2y`

=> `(dy/dx)(sin 16x + 2x*sin 2y) = (cos 2y - 16*y*cos 16x)`

=> `(dy/dx) = (cos 2y - 16*y*cos 16x)/(sin 16x + 2x*sin 2y)`

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

`dy/dx = -4`

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

=> `y - pi/4 = -4x + 2*pi`

=> `4x + y + 7*pi/4 = 0`

The tangent to `y*sin(16x) = x*cos(2y)` at `(pi/2, pi/4)` is `4x + y + 7*pi/4 = 0`

Approved by eNotes Editorial Team