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

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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 = sinx ; then dy/dx = cosx

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

Now, the given function is :-

cos(x*y) = 1 + siny

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

-sin(xy)*[y + x*(dy/dx)] = cosy*(dy/dx)

or, -y*sin(xy) -x*sin(xy)*(dy/dx) = cosy*(dy/dx)

or, -y*sin(xy) = (dy/dx)*[cosy + x*sin(xy)]

or, dy/dx = [-y*sin(xy)]/[cosy + x*sin(xy)]