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#1 2010-08-03 01:02:20

Carisma
Member
Registered: 2010-01-08
Posts: 10

Implicit Differentiation...

I have to implicitly differentiate the following and then form an equation for dy/dx:

xy/π + sinxlnx = cosx + 1

I then have to show that the gradient is lnx when x=π.

I have tried doing it several times but it seems i keep getting the wrong answer.

Any ideas?

Thanks so much in advance!

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#2 2010-08-03 05:14:44

Bob
Administrator
Registered: 2010-06-20
Posts: 10,466

Re: Implicit Differentiation...

hi Carisma

hhmmmm. me too.  I assumed that n is a constant and that sinxlnx means sinx times logbase e of x.

Here's what I did:

xy/n  + sinxlnx = cosx + 1     => y/n = [cosx + 1 –sinxlnx]/x

diff. wrt x

y/n + (x/n) .dy/dx  + (sinx)/x + cosxlnx = -sinx + 0

dy/dx = [-sinx –(sinx)/x –cosxlnx – y/n].n/x

if x = n
dy/dx = [-sinx – (sinx)/x –cosxlnx – (cosx)/x – 1/x – (sinxlnx)/x].1

I cannot see a way of making that become lnx  (the .../x and trig bits just don't cancel).

So I'm suspecting I'm mis-interpreting the problem.

Any chance you could copy the original exactly (screen shot or reference if it's off the web)?

Bob

Last edited by Bob (2010-08-03 05:17:28)


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 smile

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#3 2010-08-03 08:56:00

Bob
Administrator
Registered: 2010-06-20
Posts: 10,466

Re: Implicit Differentiation...

Are you sure the question didn't say

                    x = n times pi where n is an odd integer

       sin x is then zero and cos x = -1

then dy/dx = pi ln x.

Thought:  maybe the 'pi's have gone missing from this question.


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 smile

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