How do you apply implicit differentiation?

Question

For the equation given below, evaluate y' at the point (1, 9).

y / [ x - 7y ] = x^4 + 9

so im working on implicit differentials now and i swear it's the one concept that is brainfuckling me. i've finagled my way through a few problems, but im utterly lost on how to do this one.

Answer 1

I would put the y by itself on one side (multiply the whole thing by (x - 7y) then expand, get the terms with y and put them all one side, factor out the y and divide both sides by whats left

i got
y = (4x^5 + 9x)/(28x^4 + 64)

this should be easier to differentiate

Answer 2

Basically, just do implicit differentiation like how you do normal differentiation, except add a 'dy/dx' because you're differentiation y.

In short,
d/dx[f(y)] = dy/dx [f'(y)]

y/(x-7y) = x^4 + 9
y = (x^4 + 9)(x-7y)
d/dx(y) = d/dx[(x^4 + 9)(x-7y)]
dy/dx(1) = (x-7y)[d/dx(x^4+9)] + (x^4+9)[d/dx(x-7y)] (Chain rule)
dy/dx = (x-7y)(4x^3) + (x^4+9)(1 - dy/dx(7))
dy/dx = (x-7y)(4x^3) + (x^4 + 9 - (7x^4)dy/dx - 63(dy/dx))
(7x^4 + 63 + 1)dy/dx = 4x^4 - 4yx^3 + x^4 + 9
(7x^4 + 64)dy/dx = 5x^4 - 4yx^3 + 9
dy/dx = (5x^4 - 4yx^3 + 9) / (7x^4 + 64)

So at (1,9),

dy/dx
= (5 - 4(9) + 9) / (7 + 64)
= -22/71

Answer 3

simplify
64y=x^5+9x-7x^4y
take derivative

64y'=5x^4+9-7(4x^3y+y' x^4)
x=1
y=9
64y'=5+9-7(36+y')

71y'=14-252
71y' = 238

y'=238/71

y'=3,35


hint:
(y^3)'= 3y^2.y'
(x^4)'=4x^3