What is the tangent of (2x-1)^2 ?

The tangent of (2x-1)^2 is y=4(2a_{0}-1)x-4a_{0}^2+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of (2x-1)^2

tangent of

(2 x - 1)^{2}

y = 4 (2 a_{0} - 1) x - 4 a_{0}^{2} + 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, (2 a_{0} - 1)^{2})

Find the slope of

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

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 = 4 (2 a_{0} - 1)

Find the line with slope m=

4 (2 a_{0} - 1)

and passing through

(a_{0}, (2 a_{0} - 1)^{2}) : y = 4 (2 a_{0} - 1) x - 4 a_{0}^{2} + 1

y = 4 (2 a_{0} - 1) x - 4 a_{0}^{2} + 1

\frac{d}{d x} (ln (x^{\frac{9}{2}}))

\int \frac{16}{( x ^{2} - 4 ) ( x - 2 )} d x

\frac{( ( \frac{d y}{d x} ) - e ^{- x} + 5 )}{y} = - 5

\frac{d}{d x} (x^{0.3})

\int 2 cos^{2} (3 x) d x

What is the tangent of (2x-1)^2 ?
The tangent of (2x-1)^2 is y=4(2a_{0}-1)x-4a_{0}^2+1