What is the tangent of y=(|x|)/(sqrt(2-x^2)),(1,1) ?

The tangent of y=(|x|)/(sqrt(2-x^2)),(1,1) is y=2x-1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=(|x|)/(sqrt(2-x^2)),(1,1)

tangent of

y = \frac{∣ x ∣}{2 - x ^{2}}, (1, 1)

y = 2 x - 1

Solution steps

Find the slope of

y = \frac{∣ x ∣}{2 - x ^{2}} : \frac{d y}{d x} = \frac{2 x}{∣ x ∣ ( 2 - x ^{2} ) - x ^{2} + 2}

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 = 2

Find the line with slope m=

2

and passing through

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

y = 2 x - 1

f (x) = 5 x^{3} sin (x)

\frac{d}{d x} (ln (8 x^{3} - x^{2}))

\frac{d y}{e ^{4 y}} = e^{5 t} d t

y = 8 x^{2} + 8, (2, 40)

\int (1 + x) e^{x} d x

What is the tangent of y=(|x|)/(sqrt(2-x^2)),(1,1) ?
The tangent of y=(|x|)/(sqrt(2-x^2)),(1,1) is y=2x-1