What is the tangent of f(x)=1-x^2,(1,0) ?

The tangent of f(x)=1-x^2,(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 f(x)=1-x^2,(1,0)

tangent of

f (x) = 1 - x^{2}, (1, 0)

y = - 2 x + 2

Solution steps

Find the slope of

f (x) = 1 - x^{2} : \frac{df ( x )}{d x} = - 2 x

Compute the slope of

f (x) = 1 - x^{2}

(1, 0) : m = - 2

Find the line with slope m=

- 2

and passing through

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

y = - 2 x + 2

\frac{d}{d x} (\frac{( 6 x ^{2} + 6 x + 8 )}{x})

\frac{\partial}{\partial y} (x 1 + y^{3})

4 t \cdot sin (3 t)

\frac{x y ^{^{'}}}{y ^{2}} sin (\frac{x}{y}) = \frac{1}{y} sin (\frac{x}{y})

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

What is the tangent of f(x)=1-x^2,(1,0) ?
The tangent of f(x)=1-x^2,(1,0) is y=-2x+2