implicitly differentiate f(x)
2xy= y ^2
find second derivative with relevance to x and y.
i get y''= -xy' - y/ (y-x)^2
this is wrong though. can someone please help me. thank you
2xy' + 2y = 2y y'
2xy' - 2y y' = -2y
y'(2x - 2y) = -2y
y' = y/(y-x)
y'' = ( (y-x)y' - y(y' - 1) )/(y-x)^2
= [ (y-x)(y/(y-x) ) - y( y/(y-x) - 1) ]/(y-x)^2
= [ y - y^2 /(y-x) + y ]/(y-x)^2
= [ y - y^2 /(y-x) + y ]/(y-x)^2 * (y-x)/y-x)
=[ 2y(y-x) - y^2 ]/(y-x)^3
= (y^2 - 2x)/(y-x)^3
you have to eliminate the y' in your y'' to get only x's and y's
check my algebra, I did not write it out on paper first.
Thank you!!!! i got it :)
I didn't realize I had to omit the y' and I got y(-2x + y)/ (y-x)^3
To find the second derivative of the given equation with respect to both x and y, we can proceed as follows:
Step 1: Implicitly differentiate the equation with respect to x.
Before we begin, let's differentiate each term separately using the product rule and chain rule.
The left-hand side of the equation, 2xy, can be separated as follows:
d(2xy)/dx = 2x(dy/dx) + 2y
Now let's differentiate the right-hand side of the equation, y^2:
d(y^2)/dx = 2y(dy/dx)
So, the differentiated equation becomes:
2x(dy/dx) + 2y = 2y(dy/dx)
Simplifying the equation, we get:
2x(dy/dx) = 0
Step 2: Now, let's find the first derivative of y with respect to x (dy/dx). To do this, we need to solve the equation we obtained from Step 1 for dy/dx.
2x(dy/dx) = 0
Dividing both sides by 2x:
dy/dx = 0/(2x)
dy/dx = 0
Step 3: Up to this point, we have found the first derivative of y with respect to x. To find the second derivative, we need to differentiate the expression dy/dx = 0 with respect to x.
Differentiating y' = 0 with respect to x, we get:
d(dy/dx)/dx = d(0)/dx
d^2y/dx^2 = 0
Therefore, the second derivative (d^2y/dx^2) of the given equation is equal to 0.
Note: It seems there was a mistake in your attempt to find the second derivative. The correct answer is d^2y/dx^2 = 0, rather than -xy' - y/(y-x)^2.
Keep in mind that when implicitly differentiating an equation, you should differentiate each term separately and carefully apply the product rule and chain rule if necessary.