How do I differentiate 9cos(x)*sin(y)=5 implicitly? I am solving for dy/dx.?

Question

I really can't figure this out...

Answer 1

and niether can i

Answer 2

It's a combination of product rule and (since implicit differentiation assumes y is a function of x) chain rule.

d(9cos x * sin y)/dx = 0
d(9cos x)/dx * sin y + 9cos x * d(sin y)/dx = 0
-9sin x * sin y + 9cos x * cos y * dy/dx = 0
9 cos x * cos y * dy/dx = 9 sin x * sin y
dy/dx = (9 sin x * sin y)/(9 cos x * cos y)
dy/dx = tan x * tan y

Answer 3

9cos(x)*sin(y)=5

Product rule gives you:

-9sin(x)*sin(y) + 9cos(x)*-cos(y)*dy/dx = 0

9cos(x)*-cos(y)*dy/dx = 9sin(x)*sin(y)

dy/dx = sin(x)*sin(y) / -cos(x)*cos(y) --- don't forget the negative that's on the bottom!

And I think that is as far as you can go. I don't know of any formulas that make that look nicer...

Answer 4

You expand by taking z = 9*cos(x)*sin(y). Then you differentiate function z: dz = (dz/dx)*dx + (dz/dy)*dy. In your case dz = -9*sin(x)*sin(y)*dx + 9*coc(x)*cos(y)*dy. Since your function z = 5 is a constant dz = 0. Therefore, -9*sin(x)*sin(y)*dx + 9*cos(x)*cos(y)*dy = 0 and dy/dx = tang(x)*tang(y).

Answer 5

Let me try...

9*cos(x)*sin(y)=5 -> cos(x)*sin(y)=5/9
cos(x)*cos(y)*dy/dx + (-1)sin(x)* sin(y)=0
dy/dx= tan(x)* sin(y)/cos(y)
since
sin(y)=(5/9) sec(x)
cos(y)=(1-(25/81)sec(x)^2)^0.5
therefore
dy/dx= tan(x)*5*sec(x) / (81-25*sec(x)^2)^0.5

Hopefully this is right.

Answer 6

Yes solve for dy/dx

so cos(x)sin(y) - 5/9 = 0

sin(x)*sin(y)/dx-cos(x)cos(y)/dy = 0

sin(x)sin(y)/cos(x)cos(y) = dy/dx

tan(x)*tan(y) = dy/dx