What is the tangent of 3x^2+2 ?

The tangent of 3x^2+2 is y=6a_{0}x-3a_{0}^2+2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of 3x^2+2

tangent of

3 x^{2} + 2

y = 6 a_{0} x - 3 a_{0}^{2} + 2

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3 a_{0}^{2} + 2)

Find the slope of

y = 3 x^{2} + 2 : \frac{d y}{d x} = 6 x

Compute the slope of

y = 3 x^{2} + 2

x = a_{0} : m = 6 a_{0}

Find the line with slope m=

6 a_{0}

and passing through

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

y = 6 a_{0} x - 3 a_{0}^{2} + 2

y = (x^{2} + 3 x) 2

\frac{\partial}{\partial x} (5 x^{2} + y^{2} + 9)

\int (cos^{2} (θ))^{2} d θ

\frac{\partial}{\partial x} (e^{- 6 t} (x + 3 t))

\frac{d ^{2}}{d x ^{2}} (\frac{4 x ^{3} + 6}{x})