Find the derivative of sin(2x + 3y)= 3xy + 5y-2.
I have this so far
1) cos(2x+3y) * d/dx (2x+3y)= 3xy' + 3y + 5y'
2) cos (2x+3y) * [2 + 3y']= 3xy' + 3y + 5y'
3) 3y' cos (2x+3y) + 2cos(2x+3y)= 3xy' + 3y + 5y'
4) 3xy' + 5y' - (3y(cos(3x+3y)= 2 cos (2x + 3y) - 3y
If I am wrong, can you please explain to me what I am doing wrong. Thank you
Aside from the typos in (4) you are ok.
4) 3xy' + 5y' - 3cos(2x+3y)y' = 2 cos (2x + 3y) - 3y
5) (3x+5y-3cos(2x+3y))y' = 2cos(2x+3y) - 3y
6) y' = (2cos(2x+3y)-3y)/(3x+5y-3cos(2x+3y))
Thank you SO much sir. I feel a lot better knowing that I did this right
To find the derivative of the equation sin(2x + 3y) = 3xy + 5y - 2, you need to use implicit differentiation.
1) Start by differentiating both sides of the equation with respect to x. This means taking the derivative of sin(2x + 3y) and the derivative of each term on the right side.
d/dx [sin(2x + 3y)] = d/dx [3xy + 5y - 2]
2) Applying the chain rule on the left side, we have:
cos(2x + 3y) * (2 + 3y') = 3y + 3xy' + 5y'
Here, we are treating y as a function of x, so y' represents dy/dx, the derivative of y with respect to x.
3) Now, solve the equation for y':
Rearranging the terms, we have:
2cos(2x + 3y) + 3y'cos(2x + 3y) = 3y + 3xy' + 5y'
4) Grouping the terms with y' and moving the other terms to one side, we get:
3xy' - 3y'cos(2x + 3y) - 5y' = 3y + 2cos(2x + 3y) - 2
5) Finally, factor out y' from the left side:
y'(3x - 3cos(2x + 3y) - 5) = 3y + 2cos(2x + 3y) - 2
6) Solve for y' by dividing both sides by (3x - 3cos(2x + 3y) - 5):
y' = (3y + 2cos(2x + 3y) - 2) / (3x - 3cos(2x + 3y) - 5)
Therefore, the derivative of sin(2x + 3y) = 3xy + 5y - 2 is y' = (3y + 2cos(2x + 3y) - 2) / (3x - 3cos(2x + 3y) - 5).