What is the tangent of y=x^3 ?

The tangent of y=x^3 is y=3a_{0}^2x-2a_{0}^3

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tangent of y=x^3

tangent of

y = x^{3}

y = 3 a_{0}^{2} x - 2 a_{0}^{3}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3})

Find the slope of

y = x^{3} : \frac{d y}{d x} = 3 x^{2}

Compute the slope of

y = x^{3}

x = a_{0} : m = 3 a_{0}^{2}

Find the line with slope m=

3 a_{0}^{2}

and passing through

(a_{0}, a_{0}^{3}) : y = 3 a_{0}^{2} x - 2 a_{0}^{3}

y = 3 a_{0}^{2} x - 2 a_{0}^{3}

e^{x} sin (x)

(- 4, \frac{3 π}{4})

f (x) = \frac{1}{x}

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

y = 5 x - 1