What is the tangent of y^6+x^3=y^2+12x,(0,1) ?

The tangent of y^6+x^3=y^2+12x,(0,1) is y=3x+1

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tangent of y^6+x^3=y^2+12x,(0,1)

tangent of

y^{6} + x^{3} = y^{2} + 12 x, (0, 1)

y = 3 x + 1

Solution steps

Find the slope of

y^{6} + x^{3} = y^{2} + 12 x : \frac{d y}{d x} = \frac{12 - 3 x ^{2}}{2 y ( - 1 + 3 y ^{4} )}

Compute the slope of

y^{6} + x^{3} = y^{2} + 12 x

(0, 1) : m = 3

Find the line with slope m=

3

and passing through

(0, 1) : y = 3 x + 1

y = 3 x + 1

x \to - 1 lim (\frac{x}{0})

f (x) = 5 \cdot 3^{x}, at x = 2

\frac{d}{d x} (x^{x} + 1)

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

\int \frac{3 x ( 8 - x )}{135} d x

What is the tangent of y^6+x^3=y^2+12x,(0,1) ?
The tangent of y^6+x^3=y^2+12x,(0,1) is y=3x+1