Help with Finding the 1st and 2nd Derivative of an implicit function

Ted_Grendy

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Nov 11, 2018
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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 )
 
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. ..... Why?

To find the 2st derivative I do: -

9) I apply the Quotient rule which states :- v(du/dx) - u(dv/dx) / v^2 ...... Is that a negative sign or a "dash"?

You need parentheses to indicate grouping,

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 )
.
 


Hi Subhotosh

The reason why I don't want to factor out a 2 in step 8 is because it helps me understand the process better. Once I am confident in the process then I'll simplify as I go along.

It is a dash not a negative.

Thanks
 
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 )

The trouble is, when you use implicit differentiation, the form you get depends heavily on the particular steps you take, so it can be very hard to compare. This is because the result assumes that x and y satisfy the original equation, and ultimately you might need to express it all in terms of x, which may require solving the original equation and substituting.

Sometimes your answer is graded based on following particular steps you were taught, to avoid having to prove your answer is equivalent. So it may be counted incorrect even if it is equivalent; and equivalence is sometimes almost impossible to prove.

I took a very different approach for the second derivative (just differentiating the original equation a second time and solving for y"), and got a very simple form.

But the way to check for equivalence is to simplify what you have, and see if it matches what I got or what your teacher got. And that starts with being willing to cancel those common factors you chose to leave as they were! So give it a try, and show us your work in simplifying. (I've inserted essential parentheses to make your work mean what you intend. It does look like correct work, as far as I can see.)
 
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 )
Do not use dashes in math notation: they will confuse everybody.

Simplification reduces errors. Take advantage of it every step of the way.

There is no need to replace y' in your equation for y'': it is a rich opportunity for making the notation complicated and consequently introducing errors.

Assuming that \(\displaystyle y' = \dfrac{2 - 2x}{2y - 6}\)

and assuming that you like risking errors so you refuse to simplify, we go

\(\displaystyle u = 2 - 2x \implies u' = - \ 2.\)

\(\displaystyle v = 2y - 6 \implies v' = 2y' \text { and } y' = \dfrac{u}{v} \implies\)

\(\displaystyle y'' = \dfrac{vu' - uv'}{v^2} = \dfrac{(2y - 6)(-\ 2) - (2 - 2x)(2y')}{(2y - 6)^2} =\)

\(\displaystyle \dfrac{12 - 4y - (4y' - 4xy')}{(2y - 6)^2} = \dfrac{12 - 4y - 4y' + 4xy'}{(2y - 6)^2}.\)

Now this can be simplified to \(\displaystyle y'' = \dfrac{3 - y - y' +xy'}{(y - 3)^2}.\)

However, because you refused to simplify y', you are very likely to miss this simplification.

In any case, your step 15 is wrong because you somehow get that

\(\displaystyle y'' = -\ 2(2y - x) - \dfrac{4y' - 4xy'}{(2y - 6)^2}\),

which is obviously wrong, and therefore every thing after that is not only unnecessary but wrong too.

EDIT: I see that Dr. P has seen why you got step 15 wrong: you just missed a necessary set of parentheses, and then seem to have missed another necessary set in step 16. Easy mistake to make when the notation becomes complicated, but it will kill you on exams. A good reason to keep notation simple.
 
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original poster said:
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

I changed your equal signs to arrows. Those are not equal to
each other, despite the word "derivative" immediately above them.
 
Jeff

The only thing that Dr P added was the brackets, but I still get the result the same result i.e.:-


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 )
 
Jeff

The only thing that Dr P added was the brackets, but I still get the result the same result i.e.:-


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 affirmed that, with the corrections I made to the notation, it appeared to be correct work.

It's hard to say definitively without being able to compare it to another answer, such as my own. And that requires simplifying it.

I could just do that for you, but it's important for you to get that practice. (I often say that calculus is where students finally learn to do algebra.)

Have you tried simplifying yet? If not, why not? Simplifying is a major part of checking that it is really right, so you should be eager to do it.

Another important issue that I alluded to is, what form is required of you? Can it be left in terms of x, y, and y', or must it be in terms of x and y only? Is it supposed to be simplified? Were you told how to find second derivatives, and is that one process all you are allowed to do? (Some teachers say not to simplify, so they can expect everyone to give answers in the same form. making it easier to check.)

Also, who told you it was incorrect, and what else did they say? (Do you have a "correct" answer to compare? Did they say what was wrong with it? Do you trust them?)
 
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 )
The equal sign is one of the 1st symbols you learned about and you use it so very poorly. I was feeling ill reading what you wrote.
Namely:
1) x^2 = 2x
2) y^2 = 2yy'
3) -2x = -2
4) -6y = -6y'
5) 5 = 0
6) 5 = 0
Not one of those equals signs above are valid. 5=0??? x^2 = 2x???
Come on, you are better than that!!
 
I second wholeheartedly what Dr. Peterson said in his most recent post.

For all I know, you were originally told that it was wrong because of the missing brackets if for no other reason.

And I cannot reiterate enough that simplifying is important. It prevents mistakes; it permits "seeing" things, and it avoids complex answers being considered wrong because they differ from simpler but equivalent expressions. Finally, you may find yourself called upon to explain your results, in which case life is easier the more simply expressed are your results.

Jeff

The only thing that Dr P added was the brackets, but I still get the result the same result i.e.:-


To find the 2st derivative I do: -

15) If I put all this together I get: [-2(2y-6)-(4y' - 4xy')] / (2y-6)^2
\(\displaystyle \dfrac{-\ 2(2y - 6) - (4y' - 4xy)}{(2y - 6)^2} = \dfrac{4(3 - y - y' + xy')}{[2(y - 3)]^2} = \dfrac{3 - y - y' + xy'}{(y - 3)^2}.\)

So your answer is equivalent to, although more complex than, my previous answer (see my prior post).

Frankly, I would leave it in my simple form, but you may (absurdly) be required to give an answer in terms of x and y.

In that case, I would proceed as follows:

\(\displaystyle y'' = \dfrac{3 - y}{(y - 3)^2} + \dfrac{-\ \dfrac{1 - x}{y - 3} + \dfrac{x(1 - x)}{y -3}}{(y - 3)^2} = \dfrac{(3 - y)(y - 3)}{(y - 3)^3} + \dfrac{x - 1 + x - x^2}{(y - 3)^3} =\)

\(\displaystyle \dfrac{6y - y^2 - 9 - 1 + 2x - x^2}{(y - 3)^2} = \dfrac{x(2 - x) + y(6 - y) - 10}{(y - 3)^3}.\)

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?
When you started this thread, you made two mistakes with brackets, so that original answer was wrong. It may be as simple as that. Or it may be that you did not make the mistake with brackets in what you were told was wrong, and you had not simplified that expression. If I were the one grading, I might mistakenly mark it wrong because I did not recognize that it was equivalent to a simpler equivalent, or I might take half off for failing to simplify to the extent possible.

I end up where Dr. Peterson did: complete your answer by simplifying it.
 
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I just went through the simplifying process, and found that it did end up at my answer, in terms of x and y.

As an encouragement to do it, let me reinforce something I said earlier: the final answer is very simple, and the final step is fun! You will find that the LHS of the original equation appears in the next to last step, so you can just replace that with the RHS.

(This also serves as a reason to put it in terms of x and y, because it reveals much more about the equation than leaving y' there.)
 
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 )
Personally I would complete the squares of the original equation. I know, after proving it, that if x^2 + y^2 = r^2, then y' = -x/y. So in your case. we get y' = -(x-1)/(y-3). Taking the derivative of this, to get y", is not too bad.

What do others think about completing the square? I do realize that help was giving using the OP method.
 
Personally I would complete the squares of the original equation. I know, after proving it, that if x^2 + y^2 = r^2, then y' = -x/y. So in your case. we get y' = -(x-1)/(y-3). Taking the derivative of this, to get y", is not too bad.

What do others think about completing the square? I do realize that help was giving using the OP method.
Actually, I did think of completing the squares, but it did not seem as though it would be responsive to the OP.
 
Personally I would complete the squares of the original equation. I know, after proving it, that if x^2 + y^2 = r^2, then y' = -x/y. So in your case. we get y' = -(x-1)/(y-3). Taking the derivative of this, to get y", is not too bad.

What do others think about completing the square? I do realize that help was giving using the OP method.

Since the original question was not about finding the easiest way (in fact, it was about refusing to make things easy), the appropriate answer has been to guide the OP to find out whether his work was correct, and along the way to see the value of simplifying.

But in the end it will be good to show that completing the square makes the work considerably easier (even without using your lemma), because everything is pre-simplified. It leads very nicely to the simple form of the answer, and also shows why it is so simple. It works even better if you do it my way, differentiating the equation itself twice, rather than differentiating the quotient. (I avoid the quotient rule whenever possible.)

The benefits of thinking first, before doing the heavy work, are very much worth showing - after the OP has seen personally how much work the long way takes.
 
Hi,
I knew that something was bothering about this problem. I *think* that the two helpers for this problem might have been a bit sloppy and am curious about what everyone thinks about my concerns.

1st, since it was an equation of a circle there actually might have been 0 or 1 points on it. If the radius was negative, then there would have been no points on the curve to find the derivative of and if the radius was 0, then there would have been one point and there too would be no derivative.

2ndly, if the radius was positive, then we most certainly can find the derivative. But I feel that it should be a piecewise derivative--what was found to be the derivative but only on the boundary of the circle and undefined everywhere else.

Am I being too picky??
 
Hi,
I knew that something was bothering about this problem. I *think* that the two helpers for this problem might have been a bit sloppy and am curious about what everyone thinks about my concerns.

1st, since it was an equation of a circle there actually might have been 0 or 1 points on it. If the radius was negative, then there would have been no points on the curve to find the derivative of and if the radius was 0, then there would have been one point and there too would be no derivative.

2ndly, if the radius was positive, then we most certainly can find the derivative. But I feel that it should be a piecewise derivative--what was found to be the derivative but only on the boundary of the circle and undefined everywhere else.

Am I being too picky??
I think the bones you have to pick are with the person who devised the problem.
 
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