What is the tangent of 2x^3+4x^2-5x-3 ?

The tangent of 2x^3+4x^2-5x-3 is y=(6a_{0}^2+8a_{0}-5)x-4a_{0}^3-4a_{0}^2-3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of 2x^3+4x^2-5x-3

tangent of

2 x^{3} + 4 x^{2} - 5 x - 3

y = (6 a_{0}^{2} + 8 a_{0} - 5) x - 4 a_{0}^{3} - 4 a_{0}^{2} - 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{3} + 4 a_{0}^{2} - 5 a_{0} - 3)

Find the slope of

y = 2 x^{3} + 4 x^{2} - 5 x - 3 : \frac{d y}{d x} = 6 x^{2} + 8 x - 5

Compute the slope of

y = 2 x^{3} + 4 x^{2} - 5 x - 3

x = a_{0} : m = 6 a_{0}^{2} + 8 a_{0} - 5

Find the line with slope m=

6 a_{0}^{2} + 8 a_{0} - 5

and passing through

(a_{0}, 2 a_{0}^{3} + 4 a_{0}^{2} - 5 a_{0} - 3) : y = (6 a_{0}^{2} + 8 a_{0} - 5) x - 4 a_{0}^{3} - 4 a_{0}^{2} - 3

y = (6 a_{0}^{2} + 8 a_{0} - 5) x - 4 a_{0}^{3} - 4 a_{0}^{2} - 3

\frac{d}{d x} (2 sin (5 x))

x \to 0 + lim (8 x ln (x))

F (t) = (3 t - 1)^{4} (2 t + 1)^{- 3}

t y^{^{'}} + y = t^{4} y^{3}

\int (a x^{2} + b x + c) d x

What is the tangent of 2x^3+4x^2-5x-3 ?
The tangent of 2x^3+4x^2-5x-3 is y=(6a_{0}^2+8a_{0}-5)x-4a_{0}^3-4a_{0}^2-3