What is derivative of x^4+y^4=4a^2x^3y^3 ?

The answer to derivative of x^4+y^4=4a^2x^3y^3 is (dy)/(dx)=(x^2(3a^2y^3-x))/(y^2(y-3a^2x^3))

derivative of x^4+y^4=4a^2x^3y^3

derivative of

x^{4} + y^{4} = 4 a^{2} x^{3} y^{3}

\frac{d y}{d x} = \frac{x ^{2} ( 3 a ^{2} y ^{3} - x )}{y ^{2} ( y - 3 a ^{2} x ^{3} )}

Solution steps

x^{4} + y^{4} = 4 a^{2} x^{3} y^{3}

Treat

y

y (x)

Differentiate both sides:

4 x^{3} + 4 y^{3} \frac{d y}{d x} = 4 a^{2} (3 x^{2} y^{3} + 3 y^{2} \frac{d y}{d x} x^{3})

Isolate

\frac{d y}{d x} : \frac{d y}{d x} = \frac{x ^{2} ( 3 a ^{2} y ^{3} - x )}{y ^{2} ( y - 3 a ^{2} x ^{3} )}

\frac{d y}{d x} = \frac{x ^{2} ( 3 a ^{2} y ^{3} - x )}{y ^{2} ( y - 3 a ^{2} x ^{3} )}

d er i v a t i v e x^{4} + y^{4} = x^{2} y^{2}

d er i v a t i v e s = \frac{1}{t}

d er i v a t i v e x^{2} + y^{2} x + 6 x = 1

d er i v a t i v e x = \frac{y ^{3}}{3 + y}

d er i v a t i v e x^{2} + y^{2} = 2 x y

What is derivative of x^4+y^4=4a^2x^3y^3 ?
The answer to derivative of x^4+y^4=4a^2x^3y^3 is (dy)/(dx)=(x^2(3a^2y^3-x))/(y^2(y-3a^2x^3))