What is the tangent of y=x-x^3,(1,0) ?

The tangent of y=x-x^3,(1,0) is y=-2x+2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=x-x^3,(1,0)

tangent of

y = x - x^{3}, (1, 0)

y = - 2 x + 2

Solution steps

Find the slope of

y = x - x^{3} : \frac{d y}{d x} = 1 - 3 x^{2}

Compute the slope of

y = x - x^{3}

(1, 0) : m = - 2

Find the line with slope m=

- 2

and passing through

(1, 0) : y = - 2 x + 2

y = - 2 x + 2

\int 1 + tan (x) d x

\frac{d}{d x} (5 - 5 cos (x))

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

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