To find dy/dx, we need to use implicit differentiation.
Taking the derivative of both sides with respect to x:
2xy² + 2x²yy' = -sin(x²y²)(2xy²y' + 2x²yy')
Simplifying and solving for y':
2xy² + 2x²yy' = -sin(x²y²)2xy²y' - sin(x²y²)2x²yy'
2xy² + sin(x²y²)2xy²y' = -2x²yy' - sin(x²y²)2x²yy'
(2xy² + sin(x²y²)2xy²y') = (-2x² - sin(x²y²)2x²y')y
(2xy² + sin(x²y²)2xy²y' + 2x²y^2) = 0 (assuming y ≠ 0)
Dividing both sides by 2xy² + sin(x²y²)2xy²y' + 2x²y^2:
1 = (-2x²y)/(2xy²+sin(x²y²)2xy²y' + 2x²y^2)
Simplifying:
2xy² + sin(x²y²)2xy²y' + 2x²y^2 = -2x²y
sin(x²y²)2xy²y' = -4xy² - 2x²y^2
dy/dx = (-2xy²)/(sin(x²y²)2xy² - 2x²y^2)
Given that x²y²=cos(x²y²). Find dy/dx
3 answers
well, that went wrong. Let's pick it up at the first step where it's wrong:
2xy² + 2x²yy' = -sin(x²y²)(2xy² + 2x²yy')
2xy² + 2x²yy' = -2xy²sin(x²y²) - 2x²ysin(x²y²)y'
2xy²(1+sin(x²y²))y' = -2xy²(sin(x²y²)+1)
y' = -y/x
2xy² + 2x²yy' = -sin(x²y²)(2xy² + 2x²yy')
2xy² + 2x²yy' = -2xy²sin(x²y²) - 2x²ysin(x²y²)y'
2xy²(1+sin(x²y²))y' = -2xy²(sin(x²y²)+1)
y' = -y/x
Sorry about that mistake, thank you for correcting it!
Your solution is correct. To get to the final answer, substitute x and y back into the equation:
x²y²=cos(x²y²)
x²(cos(x²y²))y²(cos(x²y²))y' = -sin(x²y²)
2xy²(cos(x²y²))(cos(x²y²))y' + x²y²(-sin(x²y²)) = 0
2xy²(cos(x²y²))²y' = sin(x²y²)
y' = sin(x²y²)/(2xy²(cos(x²y²))²)
y' = sin(x²y²)/[2xy²cos²(x²y²)]
Your solution is correct. To get to the final answer, substitute x and y back into the equation:
x²y²=cos(x²y²)
x²(cos(x²y²))y²(cos(x²y²))y' = -sin(x²y²)
2xy²(cos(x²y²))(cos(x²y²))y' + x²y²(-sin(x²y²)) = 0
2xy²(cos(x²y²))²y' = sin(x²y²)
y' = sin(x²y²)/(2xy²(cos(x²y²))²)
y' = sin(x²y²)/[2xy²cos²(x²y²)]