How to find dy/dx of 2sin(ay)cos(bx)=1 by implicit differentiation

a=4 and b=3

and I'm looking for explanation, not just answer =P Thanks!

Well, let's put the numbers in:
2 sin (4y) cos (3x) = 1
Implicit differentiation means we just differentiate both sides of the equation. All that means is that we use the chain rule a lot on the left hand side, and remember that when you get down to y, y is also a function of x and its derivative is dy/dx. Also remember that when we differentiate, we'll get 0 on the right hand side - not 1!

d/dx (2 sin (4y) cos (3x)) = d/dx (1); product rule first:
∴ 2 sin 4y d/dx (cos 3x) + 2 [d/dx (sin 4y)] cos 3x = 0
∴ 2 sin 4y (-3 sin 3x) + 2 (4 cos 4y dy/dx) cos 3x = 0
Now we need to rearrange this equation to get dy/dx on its own:
∴ -3 sin 4y sin 3x + 4 cos 4y cos 3x dy/dx = 0
∴ 4 cos 4y cos 3x dy/dx = 3 sin 4y sin 3x
∴ dy/dx = (3 sin 4y sin 3x) / (4 cos 4y cos 3x)
= (3/4) tan 4y tan 3x.

2sin(ay)cos(bx)=1
d/dx 2sin(ay)cos(bx) = 0
2sin(ay) [- b sin bx] + cos(bx) [2a cos ay dy/dx] =0
- 2b sin(ay) [sin bx] + 2a cos(bx) cos ay dy/dx =0
2a cos(bx) cos ay dy/dx =2b sin(ay) [sin bx]
a cos(bx) cos ay dy/dx =b sin(ay) [sin bx]
dy/dx =b sin(ay) sin bx / [a cos(bx) cos ay ]
dy/dx = b tan ay tan bx /a
dy/dx = 0.75* tan 4y tan 3x
answer