How do you implicitly differentiate #9=e^ysin^2x-e^xcos^2y#?

1 Answer

#y'=(e^x cos^2 y -2*e^y sin x cos x)/(e^y sin^2 x+2 e^x sin y cos y)#

Explanation:

We start from the given equation

#9=e^y sin^2 x - e^x cos^2 y#

Differentiate with respect to x both sides of the equation

#9=e^y sin^2 x - e^x cos^2 y#

#d/dx(9)=d/dx(e^y sin^2 x) - d/dx( e^x cos^2 y)#

Use the product formula #d/dx(uv)=v d/dx(u)+u d/dx(v)# for differentiating the right side of the equation

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

#0=e^y*2*sin x cos x+sin^2 x * (e^y)y' -[ e^x*2(cos^(2-1) y)(-sin y)y'+cos^2 y*(e^x)*1]#

Simplify

#0=2 e^y sin x cos x+e^y sin^2 x * y'+2e^x sin y cos y y'-e^x cos^2 y#

Solve now for #y'#

Start transposing the terms with #y'# to one side then factor out #y'#

#e^y sin^2 x * y'+2e^x sin y cos y y'=e^x cos^2 y-2 e^y sin x cos x#

#(e^y sin^2 x +2e^x sin y cos y) y'=e^x cos^2 y-2 e^y sin x cos x#

Divide both sides of the equation by the coefficient of #y'#

#((e^y sin^2 x +2e^x sin y cos y) y')/(e^y sin^2 x +2e^x sin y cos y)=(e^x cos^2 y-2 e^y sin x cos x)/(e^y sin^2 x +2e^x sin y cos y)#

#((cancel(e^y sin^2 x +2e^x sin y cos y)) y')/cancel(e^y sin^2 x +2e^x sin y cos y)=(e^x cos^2 y-2 e^y sin x cos x)/(e^y sin^2 x +2e^x sin y cos y)#

# y'=(e^x cos^2 y-2 e^y sin x cos x)/(e^y sin^2 x +2e^x sin y cos y)#

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