Ted_Grendy
New member
- Joined
- Nov 11, 2018
- Messages
- 36
Hi all
I was hoping someone could help me confirm whether the following question was correct.
The question is; Find the 1st and 2nd derivative with respect to x of the implicit function:-
x^2 + y^2 - 2x - 6y + 5= 5
To find the 1st derivative I do: -
1) x^2 = 2x
2) y^2 = 2yy'
3) -2x = -2
4) -6y = -6y'
5) 5 = 0
6) 5 = 0
7) This gives the following: 2x + 2yy' - 2 - 6y' = 0
8) Solving for y' give me: y' = (2-2x)/(2y-6) I know that I can factor out a 2 but I do not want to.
To find the 2st derivative I do: -
9) I apply the Quotient rule which states :- v(du/dx) - u(dv/dx) / v^2
10) u = 2-2x
11) v = 2y-6
12) v*du/dx = (2y-6)*-2
13) u*dv/dx = (2-2x) * 2*y' = 4y' - 4xy'
14) v^2 = (2y-6)^2
15) If I put all this together I get: -2(2y-6)-(4y' - 4xy') / (2y-6)^2
16) Because I have already calculated the y' in steps 1-8, I replace y' with (2-2x)/(2y-6)
17) -2(2y-6)-(4[(2-2x)/(2y-6)] - 4x[(2-2x)/(2y-6)]) / (2y-6)^2
So my final answer is y'' = -2(2y-6)-(4[(2-2x)/(2y-6)] - 4x[(2-2x)/(2y-6)]) / (2y-6)^2
I know that I can simplify but I wanted to confirm this was correct as I have been told that it is not.
Any ideas?
Thank you.
(y' = 1st derivative )
(y'' = 2nd derivative )
I was hoping someone could help me confirm whether the following question was correct.
The question is; Find the 1st and 2nd derivative with respect to x of the implicit function:-
x^2 + y^2 - 2x - 6y + 5= 5
To find the 1st derivative I do: -
1) x^2 = 2x
2) y^2 = 2yy'
3) -2x = -2
4) -6y = -6y'
5) 5 = 0
6) 5 = 0
7) This gives the following: 2x + 2yy' - 2 - 6y' = 0
8) Solving for y' give me: y' = (2-2x)/(2y-6) I know that I can factor out a 2 but I do not want to.
To find the 2st derivative I do: -
9) I apply the Quotient rule which states :- v(du/dx) - u(dv/dx) / v^2
10) u = 2-2x
11) v = 2y-6
12) v*du/dx = (2y-6)*-2
13) u*dv/dx = (2-2x) * 2*y' = 4y' - 4xy'
14) v^2 = (2y-6)^2
15) If I put all this together I get: -2(2y-6)-(4y' - 4xy') / (2y-6)^2
16) Because I have already calculated the y' in steps 1-8, I replace y' with (2-2x)/(2y-6)
17) -2(2y-6)-(4[(2-2x)/(2y-6)] - 4x[(2-2x)/(2y-6)]) / (2y-6)^2
So my final answer is y'' = -2(2y-6)-(4[(2-2x)/(2y-6)] - 4x[(2-2x)/(2y-6)]) / (2y-6)^2
I know that I can simplify but I wanted to confirm this was correct as I have been told that it is not.
Any ideas?
Thank you.
(y' = 1st derivative )
(y'' = 2nd derivative )