What is the tangent of y= x/(1+x^2),(2,0.4) ?

The tangent of y= x/(1+x^2),(2,0.4) is y=-3/25 x+0.64

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y= x/(1+x^2),(2,0.4)

tangent of

y = \frac{x}{1 + x ^{2}}, (2, 0.4)

y = - \frac{3}{25} x + 0.64

Solution steps

Find the slope of

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

Compute the slope of

y = \frac{x}{1 + x ^{2}}

(2, 0.4) : m = - \frac{3}{25}

Find the line with slope m=

- \frac{3}{25}

and passing through

(2, 0.4) : y = - \frac{3}{25} x + 0.64

y = - \frac{3}{25} x + 0.64

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

x \to 4 + lim (\frac{4 - x}{∣ 4 - x ∣})

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

e^{5 t s i n (2 t)}

e^{x}, x e^{x}, 0

What is the tangent of y= x/(1+x^2),(2,0.4) ?
The tangent of y= x/(1+x^2),(2,0.4) is y=-3/25 x+0.64