What is the tangent of f(x)=(x^2+15)(2x+18) ?

The tangent of f(x)=(x^2+15)(2x+18) is y=(6a_{0}^2+36a_{0}+30)x-4a_{0}^3-18a_{0}^2+270

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tangent of f(x)=(x^2+15)(2x+18)

tangent of

f (x) = (x^{2} + 15) (2 x + 18)

y = (6 a_{0}^{2} + 36 a_{0} + 30) x - 4 a_{0}^{3} - 18 a_{0}^{2} + 270

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, (a_{0}^{2} + 15) (2 a_{0} + 18))

Find the slope of

f (x) = (x^{2} + 15) (2 x + 18) : \frac{df ( x )}{d x} = 6 x^{2} + 36 x + 30

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 6 a_{0}^{2} + 36 a_{0} + 30

Find the line with slope m=

6 a_{0}^{2} + 36 a_{0} + 30

and passing through

(a_{0}, (a_{0}^{2} + 15) (2 a_{0} + 18)) : y = (6 a_{0}^{2} + 36 a_{0} + 30) x - 4 a_{0}^{3} - 18 a_{0}^{2} + 270

y = (6 a_{0}^{2} + 36 a_{0} + 30) x - 4 a_{0}^{3} - 18 a_{0}^{2} + 270

\int \frac{16}{1 + x ^{2}} d x

\frac{d y}{d x} = 2 (1 - y)

\frac{\partial}{\partial y} (\frac{- 1}{x ^{2} + y ^{2}})

\frac{\partial}{\partial x} (e^{(4 x y)})

d er i v a t i v e y = x lo g_{e} (x)

What is the tangent of f(x)=(x^2+15)(2x+18) ?
The tangent of f(x)=(x^2+15)(2x+18) is y=(6a_{0}^2+36a_{0}+30)x-4a_{0}^3-18a_{0}^2+270