If 3xy^3 - 4x = 10y^2, find y ' (x)
a) (4 + 3y^2)/(9xy^2 + 20y)
b) (4 - 3y^2)/(9xy^2 - 20y)
c) (3 + 4y^2)/(20xy^2 + 9y)
d) (3 - 4y^2)/(20xy^2 - 9y)
why don't you show us some of the steps you would need to arrive at the correct answer of b) ?
3y^3+9xy^2*y'-4£½20y*y'
y'=b)
To find y'(x), we need to take the derivative of the equation with respect to x. Let's start by differentiating each term separately.
For the first term, 3xy^3, we apply the product rule: the derivative of the first factor (3x) times the second factor unchanged (y^3), plus the first factor (3x) times the derivative of the second factor (3y^2):
d/dx (3xy^3) = 3y^3 * dx/dx + 3x * d/dx (y^3)
= 3y^3 + 3x * (3y^2 * dy/dx)
For the second term, -4x, the derivative is simply -4.
For the third term, 10y^2, we differentiate with respect to y since it does not depend on x. We use the chain rule: the derivative of the outer function (10u^2) with respect to the inner function (y) multiplied by the derivative of the inner function (dy/dx):
d/dx (10y^2) = d/du (10u^2) * du/dx
= 20y * dy/dx
Now, let's put it all together and set it equal to 0 to solve for y ' (x):
3y^3 + 3x * (3y^2 * dy/dx) - 4x + 20y * dy/dx = 0
Now we can factor out dy/dx:
dy/dx * (3x * 3y^2 + 20y) = -3y^3 + 4x
To solve for dy/dx, divide both sides of the equation by (3x * 3y^2 + 20y):
dy/dx = (-3y^3 + 4x) / (3x * 3y^2 + 20y)
Simplifying the numerator by factoring out y^2:
dy/dx = (-y^2 * (3y - 4x)) / (3x * 3y^2 + 20y)
Further simplifying by factoring out y in the denominator:
dy/dx = (-y^2 * (3y - 4x)) / (y * (3x * 3y + 20))
Canceling out the common factor of y in the numerator and denominator:
dy/dx = (-y * (3y - 4x)) / (3x * 3y + 20)
Therefore, the correct answer is a) (4 + 3y^2)/(9xy^2 + 20y).