Given that x²y²=cos(x²y²). Find dy/dx
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
To find dy/dx, we need to differentiate both sides of the equation x²y² = cos(x²y²) with respect to x using the chain rule.
Let's start by differentiating the left-hand side of the equation.
Using the chain rule, we can rewrite x²y² as (x²) * (y²):
d/dx (x²y²) = d/dx ((x²)(y²))
Now, let's differentiate each term separately. The derivative of (x²) with respect to x is 2x, and the derivative of (y²) with respect to x is (dy/dx) * 2y.
So we have:
2x * (y²) + (x²) * (d/dx (y²)) = d/dx ((x²)(y²))
Next, let's differentiate the right-hand side of the equation.
The derivative of cos(x²y²) with respect to x is -sin(x²y²)*(d/dx (x²y²)).
So we have:
2x * (y²) + (x²) * (dy/dx) * 2y = -sin(x²y²) * (dy/dx)
Now, let's rearrange the equation to solve for dy/dx:
2x * (y²) + 2x²y * (dy/dx) = -sin(x²y²) * (dy/dx)
Move the terms with (dy/dx) to one side:
2x²y * (dy/dx) + sin(x²y²) * (dy/dx) = -2x * (y²)
Factor out (dy/dx) on the left side:
(dy/dx) * (2x²y + sin(x²y²)) = -2x * (y²)
Finally, divide both sides by (2x²y + sin(x²y²)) to solve for dy/dx:
dy/dx = (-2x * (y²)) / (2x²y + sin(x²y²))
So, the derivative dy/dx is (-2x * (y²)) / (2x²y + sin(x²y²)).
To find the derivative dy/dx of the given equation, we will use the implicit differentiation method. Here's how we can do it step by step:
Step 1: Differentiate both sides of the equation with respect to x. Keep in mind that we need to apply the chain rule whenever necessary.
On the left side:
d/dx(x²y²) = 2xy² + 2x²yy'
On the right side:
d/dx(cos(x²y²)) = -sin(x²y²) * d/dx(x²y²)
Step 2: We can rewrite the right side using the chain rule.
d/dx(x²y²) = -sin(x²y²) * [d/dx(x²y²)] * 1
Step 3: Now, we can solve for dy/dx by moving all the terms involving dy/dx to one side and simplifying:
2xy² + 2x²yy' = -sin(x²y²) * [d/dx(x²y²)] * 1
2xy² + 2x²yy' = -sin(x²y²) * 1 * [d/dx(x²y²)]
2xy² + 2x²yy' = -sin(x²y²) * d/dx(x²y²)
Step 4: Divide both sides of the equation by -sin(x²y²) to solve for dy/dx:
dy/dx = (2xy² + 2x²yy') / -sin(x²y²)
Therefore, the derivative dy/dx of the given equation x²y² = cos(x²y²) is dy/dx = (2xy² + 2x²yy') / -sin(x²y²).
Note: The equation above contains the term y', which represents the partial derivative of y with respect to x. To obtain the exact value of dy/dx, you would need additional information or constraints on the variables x and y.
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)
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)]