find y' by implicit differentiation.

i don't understand, its so complicate.

1. 4cos(x-y)=3ysinx
2. 4sin(x-y) =3ysinx

4sin(x-y)*d(x-y)=3y' + 3ycosx

On the left I applied the chain rule, and on the right i used the product rule.

4sin(x-y)*(1-y')=3y' + 3ycosx

Differentiate x - y and get 1 - y'

Then do algebra to solve for y'

4sin(x-y) - y'*4sin(x-y) = 3y' + 3ycosx

3y' + y'*4sin(x-y) = 4sin(x-2) - 3ycosx

y' = [4sin(x-2)-3ycosx]/[3+4sin(x-y)]

Try the other one yourself.

ok i'll try another, thanks

oh

d/dx cos u = - sin u du/dx
You forgot the - sign

also missed on the right side :-(

4cos(x-y)=3ysinx
-4sin(x-y)(1-y') = 3sinx*y' + 3ycosx
-4sin(x-y) + 4sin(x-y)*y' = 3sinx*y' + 3ycosx
y'(4sin(x-y) - 3sinx) = 3ycosx + 4sin(x-y)

3ycosx + 4sin(x-y)
----------------------- = y'
4sin(x-y) - 3sinx

Implicit differentiation can be a bit challenging at first, but with practice, it becomes easier. Let's go through the process step by step for each equation.

1. To find y', we'll start by differentiating both sides of the equation with respect to x. Remember that we treat y as a function of x, so we'll need to use the chain rule.

Differentiating the left side: To differentiate 4cos(x-y), we apply the chain rule. The derivative of cos(u) is -sin(u), and since u = x - y, we need to multiply by the derivative of the inside function, which is 1. So, the derivative of 4cos(x-y) is -4sin(x-y).

Differentiating the right side: To differentiate 3ysinx, we use the product rule. The derivative of 3y with respect to x is 3y', and the derivative of sinx is cosx. So, the derivative of 3ysinx is 3y'cosx + 3ysinx.

So, the equation becomes -4sin(x-y) = 3y'cosx + 3ysinx.

To solve for y', isolate it on one side of the equation. Move the terms involving y' to the left side and the remaining terms to the right side:

-4sin(x-y) - 3ysinx = 3y'cosx

Finally, divide both sides by 3cosx to obtain y':

y' = (-4sin(x-y) - 3ysinx) / (3cosx)

2. The process is similar for the second equation. Start by differentiating both sides with respect to x.

Differentiating the left side: To differentiate 4sin(x-y), we apply the chain rule. The derivative of sin(u) is cos(u), and since u = x - y, we need to multiply by the derivative of the inside function, which is 1. So, the derivative of 4sin(x-y) is 4cos(x-y).

Differentiating the right side: To differentiate 3ysinx, we use the product rule. The derivative of 3y with respect to x is 3y', and the derivative of sinx is cosx. So, the derivative of 3ysinx is 3y'cosx + 3ysinx.

So, the equation becomes 4cos(x-y) = 3y'cosx + 3ysinx.

To solve for y', isolate it on one side of the equation. Move the terms involving y' to the left side and the remaining terms to the right side:

4cos(x-y) - 3ysinx = 3y'cosx

Finally, divide both sides by 3cosx to obtain y':

y' = (4cos(x-y) - 3ysinx) / (3cosx)

Remember to simplify the equation if possible by canceling out common factors or simplifying trigonometric expressions.