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# implicit differentiation!!!

find dy/dx implicit differentiation

tan^2(x^3+y^3)=xy

### 1 Answer

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- 麻辣Lv 78 years agoFavorite Answer
xy=[tan(x^3+y^3)]^2Let z=x^3+y^3 => z'=3(x^2+y^2*y')xy'+y=2tan(z)*sec(z)^2*z'={sin(z)/[cos(z)]^3}*6(x^2+y^2*y')xy'-sin(z)*6y^2*y'/[cos(z)]^3={sin(z)*6x^2/[cos(z)]^3}-y=y'[x-sin(z)*6y^2]/[cos(z)]^3y'=[{sin(z)*6x^2/[cos(z)]^3}-y]/[x-sin(z)*6y^2]/[cos(z)]^3={6x^2*sin(z)-y*[cos(z)]^3}/[x-6y^2*sin(z)]

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