x=(6-2y²)/(1+y)
differentiate with respect to y
I saw this in a question here but..I don't get how the person go that answer.
Sorry, the other answer was incorrect.
The correct answer should have been all expressed in y instead of x, since the given expression is already in y:
You can differentiate by the product rule or the quotient rule, I prefer the former:
x=(6-2y²)/(1+y)
=(6-2y^2)(1+y)^(-1)
dx/dy
=-4y(1+y)^(-1) +(6-2y^2)(-1)(1)(x+1)^(-2)
=-4y/(y+1)-(6-2y^2)/(y+1)^2
as in the other post (with x changed for y)
You can do some algebraic simplifications if you wish.
=(1+y)d/dy(6-2y^2)-(6-2y^2)d/dy(1+y)/(1+y)^2 =(1+y)(-4y)-(6-2y^2)(1)/(1+y)^2 =-4y-4y^2-6-2y^2/(1+y)^2 =-2y^2-4y-6/(1+y)^2
To differentiate the given expression with respect to y, we need to apply the rules of differentiation. Let's break down the steps:
1. Start with the expression: x = (6 - 2y²) / (1 + y)
2. To differentiate with respect to y, we treat x as a function of y and apply the quotient rule.
The quotient rule states that for a quotient of two functions u(y) and v(y), the derivative is given by:
(u'(y)v(y) - u(y)v'(y)) / (v(y))^2
Where u'(y) represents the derivative of the numerator function u(y), and v'(y) represents the derivative of the denominator function v(y).
3. Applying the quotient rule, we get:
x' = [( (6 - 2y²)'(1 + y) ) - ( (6 - 2y²)(1 + y)' )] / (1 + y)²
Now, let's differentiate each term separately:
4. Take the derivative of the numerator function:
(6 - 2y²)'
The derivative of 6 (a constant) is 0.
The derivative of -2y² can be found using the power rule, where the derivative of x^n with respect to x is given by n*x^(n-1).
Therefore, the derivative of -2y² with respect to y is -4y.
This simplifies the numerator to:
0 - 4y = -4y
5. Take the derivative of the denominator function:
(1 + y)'
The derivative of 1 (a constant) is 0.
The derivative of y with respect to y is 1.
Therefore, the derivative of (1 + y) with respect to y is 1.
This simplifies the denominator to:
1
6. Substitute the derivative of the numerator (-4y) and the derivative of the denominator (1) back into the quotient rule equation:
x' = (-4y*1 - (6 - 2y²)*1) / (1 + y)²
= (-4y - (6 - 2y²)) / (1 + y)²
= (-4y - 6 + 2y²) / (1 + y)²
So, the derivative of x with respect to y, denoted as x', is (-4y - 6 + 2y²) / (1 + y)².