What is implicit (dy)/(dx),16x^2-16xy+y^2=1 ?

The answer to implicit (dy)/(dx),16x^2-16xy+y^2=1 is (dy)/(dx)=(8(y-2x))/(y-8x)

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implicit (dy)/(dx),16x^2-16xy+y^2=1

implicit derivative

\frac{d y}{d x}, 16 x^{2} - 16 x y + y^{2} = 1

\frac{d y}{d x} = \frac{8 ( y - 2 x )}{y - 8 x}

Solution steps

16 x^{2} - 16 x y + y^{2} = 1

Treat

y

y (x)

Differentiate both sides:

32 x - 16 (y + x \frac{d y}{d x}) + 2 y \frac{d y}{d x} = 0

Isolate

\frac{d y}{d x} : \frac{d y}{d x} = \frac{8 ( y - 2 x )}{y - 8 x}

\frac{d y}{d x} = \frac{8 ( y - 2 x )}{y - 8 x}

f (x) = (x^{3} - 9 x)^{8}, (- 3, 0)

\frac{d y}{d x} + 3 y = 2 x

x^{3}, x, 0, 1

x \to 4 lim (\frac{9}{9 - ( - 3 )})

x \to 0 lim (x^{2} \cdot ln (x))

What is implicit (dy)/(dx),16x^2-16xy+y^2=1 ?
The answer to implicit (dy)/(dx),16x^2-16xy+y^2=1 is (dy)/(dx)=(8(y-2x))/(y-8x)