- #1

karush

Gold Member

MHB

- 3,269

- 5

Find the slope of the curve at the given point}

$2y^8 + 7x^5 = 3y +6x \quad (1,1)$

Separate the variables

$2y^8-3y=-7x^5+6x$

d/dx

$16y^7y'-3y'=-35x^4+6$

isolate y'

$\displaystyle y'=\frac{-35x^4+6}{16y^7-3}$

plug in (1,1)

$\displaystyle y'=\frac{-35(1)^4+6}{16(1)^7-3}=-\frac{29}{13}=m$

so equation of tangent line is

$\displaystyle y=-\frac{29}{13}(x-1)+1$

well at least the graph seemed ok

$2y^8 + 7x^5 = 3y +6x \quad (1,1)$

Separate the variables

$2y^8-3y=-7x^5+6x$

d/dx

$16y^7y'-3y'=-35x^4+6$

isolate y'

$\displaystyle y'=\frac{-35x^4+6}{16y^7-3}$

plug in (1,1)

$\displaystyle y'=\frac{-35(1)^4+6}{16(1)^7-3}=-\frac{29}{13}=m$

so equation of tangent line is

$\displaystyle y=-\frac{29}{13}(x-1)+1$

well at least the graph seemed ok