`e^y cos(x) = 1 + sin(xy)` Find `(dy/dx)` by implicit differentiation.

Textbook Question

Chapter 3, 3.5 - Problem 19 - 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 = e^x ; then dy/dx = e^x

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 = sinx ;  then dy/dx = cosx

Now, the given function is :-

(e^y)*cosx = 1 + sin(xy)

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

-(e^y)*sinx + {(e^y)*cosx}*(dy/dx) = cos(xy)*[y + x*(dy/dx)]

[{(e^y)*cosx} - x*cos(xy)]*(dy/dx) = [y*cos(xy) + (e^y)*sinx]

or, dy/dx = [y*cos(xy) + (e^y)*sinx]/[{(e^y)*cosx} - x*cos(xy)]

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