karush
Gold Member
MHB
- 3,240
- 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
Attachments
