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ioann

Member

Registered: 2019-04-23

Posts: 3

Simplify by multiplying by denominator before Implicit Differentiation

I tried to find an answer in this forum but didn't find it. Please explain me next question

During passing tests by Implicit Differentiation I run into mistake on 6th question due of attempt to simplify the equation

I tried to multiply to the denominator

and the got next equation

and then I tried to do the Implicit Differentiation

and got next equation

but seems like I could not to do so or made a mistake, because no such answer or similar in thous answers.
Yes, I resolved this equation by using the Quotient Rule, but I cannot understand why I cannot simplify firstly

Last edited by ioann (2019-04-23 07:45:57)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,149

Re: Simplify by multiplying by denominator before Implicit Differentiation

hi ioann

Welcome to the forum.

I can find no fault with your working.

The test of whether differentiation has been done correctly must surely be "Does the gradient function have the right shape when compared with the start equation?"

So I put the two functions into the MIF equation grapher and got this result:EDIT the gradient graph here is wrong.  See later post.

The gradient curve is negative in the right places, zero at the right point, and positive in the right places.  Also the values increase and decrease at the right times.  This doesn't totally prove the differentiation is 100% correct but I'd say it is strong evidence.

LATER EDIT.  I thought I would try the quotient rule as well.  I got:

How does this compare with your answer?

Then I plotted these and the gradient curve isn't anywhere near close to what is required. ??? So I need to work on this some more.

EVEN LATER EDIT:  This is weird.  I've checked both results using Wolfram Alpha.  Both are confirmed and they are different!  The quotient answer isn't correct as it fails the basic "Does this look like the right gradient graph?" test.  ?????

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

ioann

Member

Registered: 2019-04-23

Posts: 3

Re: Simplify by multiplying by denominator before Implicit Differentiation

Bob, thanks for your answer and deep investigation.
I have looked you idea and for me the graphics are different a little (at least 2 zero points vs 1) and it correlates with your next investigations - they are different.
About the solving by Quotient Rule - yes, I got the same answer and it's the same as in tests the true result.

Seems like we miss something, but I have no idea what. Very interesting situation as for me.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,149

Re: Simplify by multiplying by denominator before Implicit Differentiation

LATER EDIT:

I think I have found the flaw in this post.  This is the corrected version.

The test of any differentiation process is "Does it give the correct gradient function?"  So I thought I would investigate this thoroughly.

The function is:

It can also be expressed:

and also re-arranged to make y the subject:

I have tried all three using the equation grapher and all give the same result:

So I then described the gradient function by experimenting with this graph.  The gradient function has these properties:

It tends to negative infinity as x approaches -1 from either side.
There is a local maximum at about -2.4 and a local minimum at about +0.4.
The gradient tends to 1 as x tends to infinity and negative infinity.

Here are the three versions of a possible gradient function:

By quotient rule:

By product rule:

By direct differentiation:

This has the right properties and so I claim is the correct answer.

My difficulty lies with the way I was using the equation grapher.

So I have used algebra instead and managed to show that all three are in fact equivalent.

By quotient rule:

Substituting

the quotient version becomes

By product rule:

So all three versions are, in fact, the same.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

ioann

Member

Registered: 2019-04-23

Posts: 3

Re: Simplify by multiplying by denominator before Implicit Differentiation

Sorry, Bob, I'm on vacation and cannot answer fast.
I have read your explanation and I agree with it. It's interesting we didn't saw it firstly .
Great job I very appreciate you help. Thank you.