implicit differentiation of x^3y^2-5x^2y+x=1
what is...
3x^2y^2 dx+2x^3ydy-10xydx-5x^2dy+dx=0
gather terms...
dx(3x^2y^2-10xy+1)+dy(2x^3y-5x^2)=0
check this, easy to make a typo
3x^2 y^2 + 2x^3 y y' - 10xy - 5x^2 y' + 1 = 0
y'(2x^3y - 5x^2) = -(3x^2y^2 - 10xy + 1)
y' = (10xy - 3x^2y^2 - 1)/(2x^3y - 5x^2)
To implicitly differentiate the equation x^3y^2 - 5x^2y + x = 1, we will treat x and y as variables and differentiate both sides of the equation with respect to x.
Let's start with differentiating each term on the left side of the equation one by one using the chain rule:
For the term x^3y^2, we have two factors x^3 and y^2. Applying the chain rule, we differentiate each factor separately with respect to x and then multiply the results:
d/dx (x^3y^2) = d/dx (x^3) * y^2 + x^3 * d/dx (y^2)
Differentiating x^3 with respect to x gives us 3x^2. Differentiating y^2 with respect to x involves treating y as a function of x, so we also multiply by dy/dx (d/dx y^2 = 2y * dy/dx):
d/dx (x^3y^2) = 3x^2 * y^2 + x^3 * 2y * dy/dx
Moving on to the next term, -5x^2y, we differentiate each factor separately with respect to x:
d/dx (-5x^2y) = -5 * d/dx (x^2) * y + -5x^2 * dy/dx
Differentiating x^2 with respect to x gives us 2x, so we have:
d/dx (-5x^2y) = -5 * 2x * y + -5x^2 * dy/dx
Finally, we differentiate the last term, x, which simply gives us 1 (the derivative of a constant):
d/dx (x) = 1
Now, let's put all the terms together:
3x^2 * y^2 + x^3 * 2y * dy/dx - 10xy - 5x^2 * dy/dx + 1 = 0
Simplifying the equation, we have:
3x^2 * y^2 + 2x^3 * y * dy/dx - 10xy - 5x^2 * dy/dx + 1 = 0
This is the implicit derivative of the given equation.