What is the tangent of 4x^3 ?

The tangent of 4x^3 is y=12a_{0}^2x-8a_{0}^3

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tangent of 4x^3

tangent of

4 x^{3}

y = 12 a_{0}^{2} x - 8 a_{0}^{3}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0}^{3})

Find the slope of

y = 4 x^{3} : \frac{d y}{d x} = 12 x^{2}

Compute the slope of

y = 4 x^{3}

x = a_{0} : m = 12 a_{0}^{2}

Find the line with slope m=

12 a_{0}^{2}

and passing through

(a_{0}, 4 a_{0}^{3}) : y = 12 a_{0}^{2} x - 8 a_{0}^{3}

y = 12 a_{0}^{2} x - 8 a_{0}^{3}

\frac{d y}{d x} = tan (x) + sec (x)

\frac{cos ( x )}{2 sin ( x )}

\frac{\partial}{\partial v} (4 u sin (v))

\int x x - 15 d x

\int x e^{x y} d y