How do you implicitly differentiate 2=e^ysiny-x^2y^3 2=eysinyx2y3?

1 Answer

y'=(2xy^3)/[e^y*cos y+e^y*sin y-3x^2y^2]

Explanation:

From the given equation

2=e^y sin y-x^2y^3

Differentiate both sides of the equation with respect to x

d/dx(2)=d/dx(e^y sin y-x^2y^3)

0=e^y*d/dx(sin y)+sin y*d/dx(e^y)-[x^2*d/dx(y^3)+y^3*d/dx(x^2)]

0=e^y*(cos y)y'+sin y*(e^y)*y'-[x^2(3y^2)y'+(y^3)(2x)]

0=e^y*(cos y)y'+sin y*(e^y)*y'-x^2(3y^2)y'-(y^3)(2x)

0=[e^y*(cos y)+sin y*(e^y)-x^2(3y^2)]y'-(y^3)(2x)

(y^3)(2x)=[e^y*(cos y)+sin y*(e^y)-x^2(3y^2)]y'

y'=(2xy^3)/[e^y*cos y+e^y*sin y-3x^2y^2]

God bless....I hope the explanation is useful.