Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

Question

y sin(2x) = x cos(2y)    at   (pi/2, pi/4)

Answer 1

Remember that the value of the derivative at the given point is equal to the slope of the tangent line at that point. Remember also that implicit differentiation means that you take the derivative of the whole thing, remember that y is a function of x, so you have to use the chain rule whenever you differentiate y. So, here we go.

 

d/dx[ysin(2x)] = d/dx[xcos(2y)]  We'll need the Product Rule on both sides.

 

 

y·cos(2x)·2 + y'·sin(2x) = x(-sin(2y)·2y') + 1·cos(2y)

 

A little cleaning up gives us  

 

2y·cos(2x) + y'·sin(2x) = -2xy'·sin(2y) + cos(2y)

 

Now we need to get the y' terms to the same side, and the other terms to the other side:

 

y'·sin(2x) + 2xy'·sin(2y) = cos(2y) - 2y·cos(2x)

 

Factor out y':

 

y'[sin(2x) + 2xsin(2y)] = cos(2y) - 2ycos(2x)

 

 

y' = [cos(2y) - 2ycos(2x)]/[sin(2x) + 2xsin(2y)]

 

Evaluate this at (π/2, π/4)

 

(cos(π/2) - (π/2)cos(π))/(sinπ + πsin(π/2)) = (0 - (π/2)(-1))/(0 + π(1))

 

   = (π/2)/π = 1/2

 

So the slope of the tangent line is 1/2

 

Use the Point-Slope form of the equation of a line:

 

y-y₁ = m(x-x₁)

 

y - π/4 = 1/2(x - π/2)

 

y = 1/2x - π/4 + π/4

 

So   y = 1/2 x is the equation of the tangent line at the point (π/2, π/4)