What is the tangent of f(x)=(x^2+36)(2x+21) ?

The tangent of f(x)=(x^2+36)(2x+21) is y=(6a_{0}^2+42a_{0}+72)x-4a_{0}^3-21a_{0}^2+756

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tangent of f(x)=(x^2+36)(2x+21)

tangent of

f (x) = (x^{2} + 36) (2 x + 21)

y = (6 a_{0}^{2} + 42 a_{0} + 72) x - 4 a_{0}^{3} - 21 a_{0}^{2} + 756

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, (a_{0}^{2} + 36) (2 a_{0} + 21))

Find the slope of

f (x) = (x^{2} + 36) (2 x + 21) : \frac{df ( x )}{d x} = 6 x^{2} + 42 x + 72

Compute the slope of

f (x) = (x^{2} + 36) (2 x + 21)

x = a_{0} : m = 6 a_{0}^{2} + 42 a_{0} + 72

Find the line with slope m=

6 a_{0}^{2} + 42 a_{0} + 72

and passing through

(a_{0}, (a_{0}^{2} + 36) (2 a_{0} + 21)) : y = (6 a_{0}^{2} + 42 a_{0} + 72) x - 4 a_{0}^{3} - 21 a_{0}^{2} + 756

y = (6 a_{0}^{2} + 42 a_{0} + 72) x - 4 a_{0}^{3} - 21 a_{0}^{2} + 756

\frac{d y}{d x} = \frac{x}{1 - 4 x ^{2}}

\frac{d ^{3}}{d x ^{3}} ((4 x - 3)^{\frac{1}{2}})

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

x \to 1 lim (\frac{∣ x - 8 ∣}{x - 8})

\frac{d}{d x} (\frac{3 x}{1 - x})

What is the tangent of f(x)=(x^2+36)(2x+21) ?
The tangent of f(x)=(x^2+36)(2x+21) is y=(6a_{0}^2+42a_{0}+72)x-4a_{0}^3-21a_{0}^2+756