Differentiate lmplicitly x^2y^2-3xy+4xy^3=4
x^2y^2-3xy+4xy^3=4
(x^2)(2y)(dy/dx) + 2x(y^2) - 3x(dy/dx) - 3y + 4x(3y^2)(dy/dx) + 4y^3 = 0
dy/dx(2x^2 y) - 3x + 12x y^2) = 3y - 2x(y^2) - 4y^3
finish it up
Oh, implicit differentiation... my mind is spinning faster than a clown car in rush hour traffic! But fear not, my friend, for I have a joke to lighten the mood:
Why don't scientists trust atoms?
Because they make up everything!
Now, let's differentiate that equation implicitly, shall we? Grab your pencil, hold on tight, and let the laughter guide our way!
Taking the derivative of each term with respect to x, we get:
d/dx (x^2y^2) - d/dx (3xy) + d/dx (4xy^3) = d/dx (4)
Oh, the thrill of derivatives... makes my wig stand on end!
Using the product rule for the first term, the chain rule for the second term, and the power rule for the third term, we have:
2xy^2 + x^2(2y)(dy/dx) - (3y + 3x(dy/dx)) + 4y^3 + 4x(3y^2)(dy/dx) = 0
Phew! Are you still with me? Now, let's gather all the terms with dy/dx on one side:
2xy^2 + 2x^2y(dy/dx) - 3y - 3x(dy/dx) + 12xy^2(dy/dx) = 0
Oh, the algebraic circus!
Next, let's factor out the dy/dx:
(2x^2y + 12xy^2 - 3x)dy/dx = 3y - 2xy^2
Almost there, my friend! Now, divide both sides by (2x^2y + 12xy^2 - 3x):
dy/dx = (3y - 2xy^2) / (2x^2y + 12xy^2 - 3x)
Ah, implicit differentiation – a thrilling circus act only for the bravest mathematicians!
To differentiate the equation implicitly, follow these steps:
Step 1: Identify the variables in the equation. In this case, the variables are x and y.
Step 2: Differentiate both sides of the equation with respect to x. Treat y as a function of x and apply the chain rule for terms involving y.
For the given equation: x^2y^2 - 3xy + 4xy^3 = 4
Differentiating both sides of the equation with respect to x:
d/dx (x^2y^2 - 3xy + 4xy^3) = d/dx(4)
Step 3: Apply the product rule and chain rule as needed.
The derivative of x^2y^2 with respect to x can be found using the product rule:
d/dx (x^2y^2) = 2xy^2 + x^2 * (d/dx (y^2))
To find d/dx (y^2), we use the chain rule:
d/dx (y^2) = d/dy (y^2) * dy/dx = 2y * dy/dx
So, d/dx (x^2y^2) = 2xy^2 + 2x^2y * dy/dx
Next, differentiate -3xy with respect to x:
d/dx (-3xy) = -3y - 3x * dy/dx
Finally, differentiate 4xy^3 with respect to x:
d/dx (4xy^3) = 4y^3 + 12xy^2 * dy/dx
Step 4: Simplify and rearrange the terms.
Putting it all together, we have:
2xy^2 + 2x^2y * dy/dx - 3y - 3x * dy/dx + 4y^3 + 12xy^2 * dy/dx = 0
Step 5: Rearrange the equation to solve for dy/dx, the derivative of y with respect to x:
Group the terms involving dy/dx:
(2x^2y + 12xy^2) * dy/dx - (3x + 3y) = 3y - 2xy^2
Divide both sides by (2x^2y + 12xy^2) to isolate dy/dx:
dy/dx = (3y - 2xy^2) / (2x^2y + 12xy^2)
This is the derivative of y with respect to x, obtained using implicit differentiation for the given equation.
To differentiate the given equation implicitly, we will consider each variable as a function of another variable and differentiate both sides of the equation with respect to that variable.
Let's consider x and y as functions of some variable, such as t. So, we have x = x(t) and y = y(t).
Differentiating both sides of the equation with respect to t, we get:
d/dt(x^2y^2) - d/dt(3xy) + d/dt(4xy^3) = d/dt(4)
To evaluate these derivatives, we need to use the chain rule and product rule.
1. Differentiating x^2y^2 with respect to t:
Using the product rule, we have:
d/dt(x^2y^2) = d/dx(x^2y^2) * dx/dt + d/dy(x^2y^2) * dy/dt
Now, d/dx(x^2y^2) = 2xy^2 and d/dy(x^2y^2) = 2x^2y. Also, dx/dt is the derivative of x with respect to t, which we'll denote as dx/dt = x'. Similarly, dy/dt = y'.
Therefore, d/dt(x^2y^2) = 2xy^2 * x' + 2x^2y * y'.
2. Differentiating 3xy with respect to t:
Using the product rule, we have:
d/dt(3xy) = d/dx(3xy) * dx/dt + d/dy(3xy) * dy/dt
Now, d/dx(3xy) = 3y and d/dy(3xy) = 3x.
Therefore, d/dt(3xy) = 3y * x' + 3x * y'.
3. Differentiating 4xy^3 with respect to t:
Using the product rule, we have:
d/dt(4xy^3) = d/dx(4xy^3) * dx/dt + d/dy(4xy^3) * dy/dt
Now, d/dx(4xy^3) = 4y^3 and d/dy(4xy^3) = 12xy^2.
Therefore, d/dt(4xy^3) = 4y^3 * x' + 12xy^2 * y'.
4. Differentiating 4 with respect to t:
The derivative of a constant is 0. Therefore, d/dt(4) = 0.
Substituting these derivatives back into the equation, we have:
2xy^2 * x' + 2x^2y * y' - (3y * x' + 3x * y') + (4y^3 * x' + 12xy^2 * y') = 0
Now, we can simplify and collect the terms with x' and y' on one side:
2xy^2 * x' - 3y * x' + 4y^3 * x' = (3x * y' - 2x^2y * y' - 12xy^2 * y')
Factoring out x' on the left side and y' on the right side:
x'(2xy^2 - 3y + 4y^3) = y'(3x - 2x^2y - 12xy^2)
Finally, we can solve for x' and y' by dividing both sides by the respective factors:
x' = (3x - 2x^2y - 12xy^2) / (2xy^2 - 3y + 4y^3)
y' = (2xy^2 - 3y + 4y^3) / (3x - 2x^2y - 12xy^2)
These are the equations for the derivatives of x and y implicitly defined by the given equation.