What is the tangent of f(x)=(x^3)/3+6x^2+36x+9 ?

The tangent of f(x)=(x^3)/3+6x^2+36x+9 is y=(a_{0}^2+12a_{0}+36)x-a_{0}^3-6a_{0}^2+(a_{0}^3)/3+9

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tangent of f(x)=(x^3)/3+6x^2+36x+9

tangent of

f (x) = \frac{x ^{3}}{3} + 6 x^{2} + 36 x + 9

y = (a_{0}^{2} + 12 a_{0} + 36) x - a_{0}^{3} - 6 a_{0}^{2} + \frac{a _{0}^{3}}{3} + 9

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{a _{0}^{3}}{3} + 6 a_{0}^{2} + 36 a_{0} + 9)

Find the slope of

f (x) = \frac{x ^{3}}{3} + 6 x^{2} + 36 x + 9 : \frac{df ( x )}{d x} = x^{2} + 12 x + 36

Compute the slope of

f (x) = \frac{x ^{3}}{3} + 6 x^{2} + 36 x + 9

x = a_{0} : m = a_{0}^{2} + 12 a_{0} + 36

Find the line with slope m=

a_{0}^{2} + 12 a_{0} + 36

and passing through

(a_{0}, \frac{a _{0}^{3}}{3} + 6 a_{0}^{2} + 36 a_{0} + 9) : y = (a_{0}^{2} + 12 a_{0} + 36) x - a_{0}^{3} - 6 a_{0}^{2} + \frac{a _{0}^{3}}{3} + 9

y = (a_{0}^{2} + 12 a_{0} + 36) x - a_{0}^{3} - 6 a_{0}^{2} + \frac{a _{0}^{3}}{3} + 9

\int \frac{x}{1 + 2 x} d x

\frac{d}{d x} (- (x + 3) (x - 9)^{2})

\int x \cdot e^{x y} d y

\int 16 x e^{4 x} d x

x^{2} + 16

What is the tangent of f(x)=(x^3)/3+6x^2+36x+9 ?
The tangent of f(x)=(x^3)/3+6x^2+36x+9 is y=(a_{0}^2+12a_{0}+36)x-a_{0}^3-6a_{0}^2+(a_{0}^3)/3+9