What is the tangent of y=x^3+3x^2 ?

The tangent of y=x^3+3x^2 is y=(3a_{0}^2+6a_{0})x-2a_{0}^3-3a_{0}^2

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tangent of y=x^3+3x^2

tangent of

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

y = (3 a_{0}^{2} + 6 a_{0}) x - 2 a_{0}^{3} - 3 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} + 3 a_{0}^{2})

Find the slope of

y = x^{3} + 3 x^{2} : \frac{d y}{d x} = 3 x^{2} + 6 x

Compute the slope of

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

x = a_{0} : m = 3 a_{0}^{2} + 6 a_{0}

Find the line with slope m=

3 a_{0}^{2} + 6 a_{0}

and passing through

(a_{0}, a_{0}^{3} + 3 a_{0}^{2}) : y = (3 a_{0}^{2} + 6 a_{0}) x - 2 a_{0}^{3} - 3 a_{0}^{2}

y = (3 a_{0}^{2} + 6 a_{0}) x - 2 a_{0}^{3} - 3 a_{0}^{2}

\frac{\partial}{\partial x} ((x^{5} + x - y)^{6})

y = 9 e^{- x} + e^{2 x}

\int 8 x^{3} - cos (4 x) d x

e

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

What is the tangent of y=x^3+3x^2 ?
The tangent of y=x^3+3x^2 is y=(3a_{0}^2+6a_{0})x-2a_{0}^3-3a_{0}^2