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

The tangent of y=x^3,(1,0) is y=0 \lor y= 27/4 x-27/4

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tangent of y=x^3,(1,0)

tangent of

y = x^{3}, (1, 0)

y = 0 or y = \frac{27}{4} x - \frac{27}{4}

Solution steps

Slope of the tangent line at

(1, 0) : m = \frac{x ^{3}}{x - 1}

Find the slope of

y = x^{3} : m = 3 x^{2}

Solve

3 x^{2} = \frac{x ^{3}}{x - 1} : x = 0, x = \frac{3}{2}

Find the tangent line at

x = 0 : y = 0

Find the tangent line at

x = \frac{3}{2} : y = \frac{27}{4} x - \frac{27}{4}

y = 0 or y = \frac{27}{4} x - \frac{27}{4}

\frac{d}{d x} (sin (3 x + y z))

\frac{d}{d x} (x^{0.7})

\int_{0}^{2 π} t^{2} sin (2) d t

\frac{\partial}{\partial v} (sin (u) sin (v))

x \to - 2 lim (2 x^{2} + 5 x + 3)

What is the tangent of y=x^3,(1,0) ?
The tangent of y=x^3,(1,0) is y=0 \lor y= 27/4 x-27/4