`y cos(x) = x^2 + y^2` Find `(dy/dx)` by implicit differentiation.

Textbook Question

Chapter 3, 3.5 - Problem 11 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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hkj1385 | (Level 1) Assistant Educator

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

2) If y = x^2 ; then dy/dx = 2x

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

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

Now, the given function is :-

y*cosx = (x^2) + (y^2)

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

-(y*sinx) + cosx*(dy/dx) = 2x + 2y*(dy/dx)

or, [cosx - 2y]*(dy/dx) = (2x + y*sinx)

or, dy/dx = (2x + y*sinx)/[cosx - 2y]

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