What is the tangent of y= 7/(x^2+3),(2,1) ?

The tangent of y= 7/(x^2+3),(2,1) is y=-4/7 x+15/7

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y= 7/(x^2+3),(2,1)

tangent of

y = \frac{7}{x ^{2} + 3}, (2, 1)

y = - \frac{4}{7} x + \frac{15}{7}

Solution steps

Find the slope of

y = \frac{7}{x ^{2} + 3} : \frac{d y}{d x} = - \frac{14 x}{( x ^{2} + 3 ) ^{2}}

Compute the slope of

y = \frac{7}{x ^{2} + 3}

(2, 1) : m = - \frac{4}{7}

Find the line with slope m=

- \frac{4}{7}

and passing through

(2, 1) : y = - \frac{4}{7} x + \frac{15}{7}

y = - \frac{4}{7} x + \frac{15}{7}

x \to 2 lim (x^{3} + 4 x^{2} - 3 x - 6)

\int \frac{1}{t ^{2} 5 - t ^{2}} d t

\int_{1}^{4} \frac{3}{x ^{2}} d x

\frac{\partial}{\partial x} (x + y + x y + \frac{x}{y} + \frac{y}{x})

y^{^{''}} + y = k

What is the tangent of y= 7/(x^2+3),(2,1) ?
The tangent of y= 7/(x^2+3),(2,1) is y=-4/7 x+15/7