What is the tangent of 7xy^3+5xy=12,\at x=1 ?

The tangent of 7xy^3+5xy=12,\at x=1 is y=-6/13 x+19/13

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of 7xy^3+5xy=12,\at x=1

tangent of

7 x y^{3} + 5 x y = 12, at x = 1

y = - \frac{6}{13} x + \frac{19}{13}

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

7 x y^{3} + 5 x y = 12 : \frac{d y}{d x} = \frac{- 7 y ^{3} - 5 y}{x ( 5 + 21 y ^{2} )}

Compute the slope of

7 x y^{3} + 5 x y = 12

x = 1 : m = - \frac{6}{13}

Find the line with slope m=

- \frac{6}{13}

and passing through

(1, 1) : y = - \frac{6}{13} x + \frac{19}{13}

y = - \frac{6}{13} x + \frac{19}{13}

(2 x + 3) + (2 y - 2) y^{^{'}} = 0

\int x^{2} x - 3 d x

\frac{d}{d x} (x^{3} - 2 x^{2} + 4)

x \to 0 lim (\frac{3 sin ( x )}{x})

\frac{1}{2 + x}

What is the tangent of 7xy^3+5xy=12,\at x=1 ?
The tangent of 7xy^3+5xy=12,\at x=1 is y=-6/13 x+19/13