What is the tangent of x^3+2x^2+4 ?

The tangent of x^3+2x^2+4 is y=(3a_{0}^2+4a_{0})x-2a_{0}^3-2a_{0}^2+4

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tangent of x^3+2x^2+4

tangent of

x^{3} + 2 x^{2} + 4

y = (3 a_{0}^{2} + 4 a_{0}) x - 2 a_{0}^{3} - 2 a_{0}^{2} + 4

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} + 2 a_{0}^{2} + 4)

Find the slope of

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

Compute the slope of

y = x^{3} + 2 x^{2} + 4

x = a_{0} : m = 3 a_{0}^{2} + 4 a_{0}

Find the line with slope m=

3 a_{0}^{2} + 4 a_{0}

and passing through

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

y = (3 a_{0}^{2} + 4 a_{0}) x - 2 a_{0}^{3} - 2 a_{0}^{2} + 4

\int \frac{4 ^{x}}{x} d x

y^{^{''}} + 9 y = 4 sin (x)

\frac{d}{d x} (1 + 5 x)

\frac{\partial}{\partial x} (3 e^{- y} (x^{2} + y^{2}) + 4)

\int \frac{1}{x ( x ^{2} - 16 ) ^{\frac{3}{2}}} d x

What is the tangent of x^3+2x^2+4 ?
The tangent of x^3+2x^2+4 is y=(3a_{0}^2+4a_{0})x-2a_{0}^3-2a_{0}^2+4