What is implicit (dy)/(dx),y^{2/3}-x^2+xy=1 ?

The answer to implicit (dy)/(dx),y^{2/3}-x^2+xy=1 is (dy)/(dx)=(6xy^{1/3}-3y^{4/3})/(2+3xy^{1/3)}

implicit (dy)/(dx),y^{2/3}-x^2+xy=1

implicit derivative

\frac{d y}{d x}, y^{\frac{2}{3}} - x^{2} + x y = 1

\frac{d y}{d x} = \frac{6 x y ^{\frac{1}{3}} - 3 y ^{\frac{4}{3}}}{2 + 3 x y ^{\frac{1}{3}}}

Solution steps

y^{\frac{2}{3}} - x^{2} + x y = 1

Treat

y

y (x)

Differentiate both sides:

\frac{2 \frac{d y}{d x} - 6 x y ^{\frac{1}{3}} + 3 y ^{\frac{4}{3}} + 3 x \frac{d y}{d x} y ^{\frac{1}{3}}}{3 y ^{\frac{1}{3}}} = 0

Isolate

\frac{d y}{d x} : \frac{d y}{d x} = \frac{6 x y ^{\frac{1}{3}} - 3 y ^{\frac{4}{3}}}{2 + 3 x y ^{\frac{1}{3}}}

\frac{d y}{d x} = \frac{6 x y ^{\frac{1}{3}} - 3 y ^{\frac{4}{3}}}{2 + 3 x y ^{\frac{1}{3}}}

\frac{\partial}{\partial x} (5 x sin (3 x^{2} y))

f (x) = x^{2} - x + 13, at x = 4

\int \frac{x ^{2}}{x ^{3}} + 5 d x

\frac{r}{r ^{2} + 3}

8 sin (t)

What is implicit (dy)/(dx),y^{2/3}-x^2+xy=1 ?
The answer to implicit (dy)/(dx),y^{2/3}-x^2+xy=1 is (dy)/(dx)=(6xy^{1/3}-3y^{4/3})/(2+3xy^{1/3)}