Find dy/dx by implicit differentiation.
x^3 - 3x^2y + 2xy^2 =12
Please show me the work/steps on how to do it.
x^3 - (3x^2)(y) + (2x)(y^2) =12
use the product rule on the 2nd and 3rd term
3x^2 - 3x^2 dy/dx - 6xy + 2x (2y)dy/dx + 2y^2 = 0
dy/dx(4xy - 3x^2) = 6xy - 3x^2 - 2y^2
dy/dx = (6xy - 3x^2 - 2y^2))/(4xy - 3x^2)
To find dy/dx by implicit differentiation, follow these steps:
1. Differentiate both sides of the equation with respect to x, treating y as a function of x.
2. Use the product rule, chain rule, and power rule as needed to differentiate each term.
3. Solve the resulting equation for dy/dx.
Now let's go through the steps to find dy/dx for the given equation x^3 - 3x^2y + 2xy^2 = 12:
1. Differentiate both sides of the equation with respect to x:
d/dx (x^3 - 3x^2y + 2xy^2) = d/dx (12)
2. Differentiate each term separately:
d/dx (x^3) - d/dx (3x^2y) + d/dx (2xy^2) = 0
3. Apply the power rule, product rule, and chain rule as needed:
3x^2 - 2(3x^2y)' + 2(xy^2)' = 0
For the first term, x^3, the derivative is simply the power rule: (3x^2).
For the second term, -3x^2y, we use the product rule. The first part of the rule is the derivative of -3x^2, which is -6x. The second part is the derivative of y, which is dy/dx. Notice that we treat y as a function of x and take its derivative. So the derivative of -3x^2y with respect to x is -3x^2(dy/dx).
For the third term, 2xy^2, we again use the product rule. The first part is the derivative of 2x, which is 2. The second part is the derivative of y^2, which we find using the chain rule. The derivative of y^2 with respect to y is 2y, and then we multiply by the derivative of y with respect to x, which is dy/dx. So overall, the derivative of 2xy^2 with respect to x is 2(dy/dx)y^2 + 2xy(2y)(dy/dx) = 2y^2(dy/dx) + 4xy^2(dy/dx).
Putting it together, the equation becomes:
3x^2 - 3x^2(dy/dx) + 2y^2(dy/dx) + 4xy^2(dy/dx) = 0
4. Now, solve the equation for dy/dx:
Bring all terms containing dy/dx to one side:
-3x^2(dy/dx) + 2y^2(dy/dx) + 4xy^2(dy/dx) = -3x^2 + 12xy^2 - 2y^2
Factor out dy/dx:
(2y^2 + 4xy^2 - 3x^2)(dy/dx) = -3x^2 + 12xy^2 - 2y^2
Divide both sides by (2y^2 + 4xy^2 - 3x^2):
dy/dx = (-3x^2 + 12xy^2 - 2y^2)/(2y^2 + 4xy^2 - 3x^2)
And there you have it! The derivative dy/dx for the given equation is (-3x^2 + 12xy^2 - 2y^2)/(2y^2 + 4xy^2 - 3x^2).