`y sin(2x) = x cos(2y), (pi/2, pi/4)` Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

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Chapter 3, 3.5 - Problem 25 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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hkj1385 | (Level 1) Assistant Educator

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Note:- 1) If  y = sinx ; then dy/dx = cosx 

2) If y = cosx ; then dy/dx = -sinx

3) If y = u*v ; where both u & v are functions of 'x' , then

dy/dx = u*(dv/dx) + v*(du/dx)

4) If y = k ; where 'k' = constant ; then dy/dx = 0

5) cos(pi/2) = sin(pi) = 0

6) sin(pi/2) = 1 & cos(pi) = -1

Now, the given function is :-

y*sin(2x) = x*cos(2y)

Differentiating both sides w.r.t 'x' we get

{sin(2x)}*(dy/dx) + 2y*cos(2x) = cos(2y) - 2x*{sin(2y)}*(dy/dx)

Putting x = pi/2 & y = pi/4 in the above equation we get

dy/dx = slope of the tangent at point (pi/2,pi/4) = 1/2

Thus, the eqaution of the tangent at the point (pi/2,pi/4) is:-

y - (pi/4) = (1/2)*{x - (pi/2)}

y = (1/2)*x = tangent equation

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