What is the tangent of 7x^2+xy+7y^2=15,(1,1) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of 7x^2+xy+7y^2=15,(1,1)

tangent of

7 x^{2} + x y + 7 y^{2} = 15, (1, 1)

y = - x + 2

Solution steps

Find the slope of

7 x^{2} + x y + 7 y^{2} = 15 : \frac{d y}{d x} = \frac{- 14 x - y}{x + 14 y}

Compute the slope of

7 x^{2} + x y + 7 y^{2} = 15

(1, 1) : m = - 1

Find the line with slope m=

- 1

and passing through

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

y = - x + 2

x \to 0 + lim ((e^{x} + x)^{\frac{1}{x}})

x \to 3 lim (\frac{x ^{2} - 3}{x - 3})

\int \frac{20 x}{1 - x ^{4}} d x

\int \frac{1}{4 + 2 x} d x

y = x^{3} + 7 x - 5, (1, 3)

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