What is the tangent of (25x^3)/3+10x^2+4x-7 ?

The tangent of (25x^3)/3+10x^2+4x-7 is y=(25a_{0}^2+20a_{0}+4)x-25a_{0}^3-10a_{0}^2+(25a_{0}^3)/3-7

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tangent of (25x^3)/3+10x^2+4x-7

tangent of

\frac{25 x ^{3}}{3} + 10 x^{2} + 4 x - 7

y = (25 a_{0}^{2} + 20 a_{0} + 4) x - 25 a_{0}^{3} - 10 a_{0}^{2} + \frac{25 a _{0}^{3}}{3} - 7

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{25 a _{0}^{3}}{3} + 10 a_{0}^{2} + 4 a_{0} - 7)

Find the slope of

y = \frac{25 x ^{3}}{3} + 10 x^{2} + 4 x - 7 : \frac{d y}{d x} = 25 x^{2} + 20 x + 4

Compute the slope of

y = \frac{25 x ^{3}}{3} + 10 x^{2} + 4 x - 7

x = a_{0} : m = 25 a_{0}^{2} + 20 a_{0} + 4

Find the line with slope m=

25 a_{0}^{2} + 20 a_{0} + 4

and passing through

(a_{0}, \frac{25 a _{0}^{3}}{3} + 10 a_{0}^{2} + 4 a_{0} - 7) : y = (25 a_{0}^{2} + 20 a_{0} + 4) x - 25 a_{0}^{3} - 10 a_{0}^{2} + \frac{25 a _{0}^{3}}{3} - 7

y = (25 a_{0}^{2} + 20 a_{0} + 4) x - 25 a_{0}^{3} - 10 a_{0}^{2} + \frac{25 a _{0}^{3}}{3} - 7

f (x) = \frac{2}{x + 3}

\frac{\partial}{\partial y} (x^{y} ln (x))

\int_{0}^{34} x^{2} - 17 d x

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

\int_{0}^{1} x^{2} + 2 d x

What is the tangent of (25x^3)/3+10x^2+4x-7 ?
The tangent of (25x^3)/3+10x^2+4x-7 is y=(25a_{0}^2+20a_{0}+4)x-25a_{0}^3-10a_{0}^2+(25a_{0}^3)/3-7