if x^2 + y^3 = (x^3)(y^2) then dy/dx = ? thanks
x^2 + y^3 = (x^3)(y^2)
differentiate with respect to x
2x +3y²(dy/dx) = 3x²y²+x³*2y*(dy/dx)
Group dy/dx on left:
(dy/dx)(3y²-2x³y)=3x²y²-2x
dy/dx=(3x²y²-2x)/(3y²-2x³y)
check my work.
implicit differentiation y^5 + x^2y^3=1 + x^4y
To find dy/dx, let's begin by differentiating both sides of the given equation with respect to x.
Differentiating x^2 + y^3 = x^3 * y^2 with respect to x:
(d/dx)(x^2) + (d/dx)(y^3) = (d/dx)(x^3 * y^2)
Using the power rule, we find:
2x + 3y^2 * (dy/dx) = 3x^2 * y^2 + 2x^3 * y * (dy/dx)
Now, let's isolate dy/dx by bringing all terms involving it to one side of the equation:
3y^2 * (dy/dx) - 2x^3 * y * (dy/dx) = 3x^2 * y^2 - 2x
Factoring out (dy/dx) on the left side of the equation:
(3y^2 - 2x^3 * y) * (dy/dx) = 3x^2 * y^2 - 2x
Finally, solving for dy/dx, divide both sides of the equation by (3y^2 - 2x^3 * y):
(dy/dx) = (3x^2 * y^2 - 2x) / (3y^2 - 2x^3 * y)
Thus, dy/dx = (3x^2 * y^2 - 2x) / (3y^2 - 2x^3 * y).