calculus question?

Question

equation of the line to the tangent curve at

x ln y+ 5xy=20 at the point 4,1

Answer 1

Differentiate implicitly,

lny + xy'/y + 5y + 5xy' = 0

y'/y (x + 5xy) = -ln y - 5y

multiply by Y and divide by (x + 5xy) on both sides,

y' = [- y ln(y) - 5y^2]/ (x + 5xy)

y'(4 ,1) = -5 /24 ..............SLOPE

Eq. of the tangent line is,

y - 1 = -5/24 (x - 4)
24y - 24 = -5x + 20

24y + 5x - 44 = 0

Answer 2

lny+ xy'/y+5y+5xy'=0

ln1+4y'/1+5.1+5.4.y'=0

0+24y'+5=0

y'=-5/24=slope

equation of the line to the tangent curve is
y-1=-5/24 . (x-4)

Answer 3

Implicit differentiation:

x(1/y)dy/dx+lny+5(xdy/dx+y)=0

(x/y)dy/dx+5xdy/dx=-(lny+5y)

dy/dx= - (lny+5y)/(5x+x/y)

dy/dx | (4,1)

= - (0+5)/(20+4)= -5/24

So the equation of the tangent line here is:

y-1= -5/24 (x-4)

Or

y= (-5/24)x+11/6