calculus ?

Question

Okay, I'm trying to differentiate the following equation with respect to x:

4sinxcosy = 1

I got an answer but I don't know if it's right: cotxcosy*cosxcoty = dy/dx

If mine's not correct, can you show me how to get the right answer? Thanks a lot!

Answer 1

4sin(x)cos(y) = 1

Before I implicitly differentiate, I'm going to divide both sides by 4.

sin(x)cos(y) = 1/4

Now, we differentiate. Use the product rule on the left hand side, and the chain rule on the "y" term inside cos(y).

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

Move the cos(x)cos(y) over to the right hand side, to get

-sin(x)sin(y) [dy/dx] = -cos(x)cos(y)

Now, divide both sides by -sin(x)sin(y), to get

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

We can turn this into a product of fractions.

dy/dx = [cos(x)/sin(x)] [cos(y)/sin(y)]
dy/dx = cot(x) cot(y)

Answer 2

sinx*cosy=0.25
Using product rule (and chain rule)
(sinx)'(cosy)+(sinx)(cosy)'=0
cosx*cosy+sinx*-siny*y'=0
Solving for y'
y'=(cosxcosy)/(sinxsiny)
OR
y'=cotx*coty

If you want it in terms of x only, solve the original equation in terms of y:

cosy=1/(4sinx)
Setup a right triangle. Cosine is Adjacent over Hypotenuese, so your right triangle has an adjacent side of 1, a hypoteneuse of 4sinx, making the opposite side sqrt(16sin^2(x)-1).

Thus, for y' you need coty:

coty=adjacent/opposite=1/sqrt(16sin^2(x)-1)
Then substitute into your answer to get your function purely a function of x:

y'=cotx/sqrt(16sin^2(x)-1)

Viola!

Answer 3

4cosxcosy -4 sinxsinydy/dx = 0
cosxcosy/sinxsiny = dy/dx
cotx * coty = dy/dx

i don't know why you have those cos's in there, but they shouldn't be there.

Answer 4

You know what? If you can, stay clear of that. It's tough subject.