My Math Forum  

Go Back   My Math Forum > High School Math Forum > Algebra

Algebra Pre-Algebra and Basic Algebra Math Forum


Reply
 
LinkBack Thread Tools Display Modes
May 13th, 2010, 05:00 PM   #1
Senior Member
 
Joined: Mar 2010
From: Melbourne

Posts: 178
Thanks: 0

1st and 2nd Derivatives ..

HI, I have:

e^(x) / x

I need to find the 1st and 2nd derivatives.

--

My first question is, when finding the first derivative, I converted it from the quotient representation to the product representation for ease.

e^(x)/x == e^(x)(x^(-1/2))

Finding the derivative of this (using the product rule), I have:

(e^(x)(x^-1)) + (e^(x)-1x^(-1/2))
= (e^(x)(x^-1)) - 1(e^(x)x^(-1/2))
= (e^(x)(x^-1)) - (e^(x)x^(-1/2))
= (e^(x)/(x)) - (e^(x)/x^(1/2))

Now, Wolfram Alpha is reporting: = (e^(x)/(x)) - (e^(x)/x^(2)) ... how does (e^(x)x^(-1/2)) become: (e^(x)/x^(2)) ??

wulfgarpro.
wulfgarpro is offline  
 
May 13th, 2010, 05:40 PM   #2
Global Moderator
 
The Chaz's Avatar
 
Joined: Nov 2009
From: Northwest Arkansas

Posts: 2,766
Thanks: 4

Re: 1st and 2nd Derivatives ..

Something's fishy...
(e^x)/x = (x^-1)(e^x)
So you can differentiate this using the product rule...

(x^-1)` *(e^x) + (x^-1)(e^x)` (I'm using ` to represent the derivative...)
(-1x^-2)(e^x) + (x^-1)(e^x)
(-1x^-2)(e^x) + x(x^-2)(e^x) (Multiply and "divide" by x to get a "common denominator", of sorts)
factor out...
(e^x)(x^-2)[-1 + x]
Got it from here?
The Chaz is offline  
May 13th, 2010, 06:59 PM   #3
Senior Member
 
Joined: Mar 2010
From: Melbourne

Posts: 178
Thanks: 0

Re: 1st and 2nd Derivatives ..

Wow .. I lost the plot. Would be great if someone could confirm my ANS:

Find the 2nd derivative of e(^x)/x.

--

e(^x)/x is equivalent to: e^(x)x^(-1)

Using the product rule:

(x^(-1)e^(x)) + (-1x^(-2)e^(x))
= e^(x)/x - e^(x)/x^(2) :: 1ST Derivative

--

Now, finding the 2nd derivative:

e^(x)/x - e^(x)/x^(2) is equivalent to: (e^(x)x^(-1)) - (e^(x)x^(-2))

Using the product rule:

(e^(x)x^(-1))

= (e^(x)x^(-1)) + (e^(x)-1x^(-2))
= (e^(x)/x) - (e^(x)/x^(2))

(e^(x)x^(-2))

= (e^(x)x^(-2)) + (e^(x)-2x^(-3))
= (e^(x)/x^(2)) - (2e^(x)/x^(3))

Adding back these derivatives back into the 1st derivative we have:

= (e^(x)/x) - (e^(x)/x^(2)) - (e^(x)/x^(2)) - (2e^(x)/x^(3))
= (e^(x)/x) - (2e^(x)/x^(2)) - (2e^(x)/x^(3)) :: 2ND Derivative

--

wulfgarpro.
wulfgarpro is offline  
May 13th, 2010, 07:06 PM   #4
Global Moderator
 
greg1313's Avatar
 
Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,958
Thanks: 1146

Math Focus: Elementary mathematics and beyond
Re: 1st and 2nd Derivatives ..

Everything looks good to me except (e^(x)/x) - (2e^(x)/x^(2)) + (2e^(x)/x^(3)).
greg1313 is offline  
May 13th, 2010, 08:39 PM   #5
Senior Member
 
Joined: Mar 2010
From: Melbourne

Posts: 178
Thanks: 0

Re: 1st and 2nd Derivatives ..

Hm,

From:

(e^x)(x^-1)+(e^x)(-1x^-2)-(e^x)(-1x^-2)+(e^x)(2x^-3)

I know the correct answer is:

{e^x} / {x} - {e^x} / {x^2} - {e^x} / {x^2} + {2e^x} / {x^3}

But from what I can see, does not -(e^x)(-1x^-2) become +(e^x)(-1x^-2) ?

wulfgarpro.
wulfgarpro is offline  
May 13th, 2010, 08:44 PM   #6
Senior Member
 
Joined: Nov 2008

Posts: 265
Thanks: 0

Re: 1st and 2nd Derivatives ..

Using the quotient rule, I get [e^x(x^2 - 2x + 2)]/x^3 for the second derivative.

In situations like this, I've always found it easier to use the quotient rule for some reason.
mathman2 is offline  
May 13th, 2010, 08:49 PM   #7
Senior Member
 
Joined: Mar 2010
From: Melbourne

Posts: 178
Thanks: 0

Re: 1st and 2nd Derivatives ..

I'm so close though. One sign out. 0.o.

wulfgarpro.
wulfgarpro is offline  
May 13th, 2010, 09:15 PM   #8
Global Moderator
 
greg1313's Avatar
 
Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,958
Thanks: 1146

Math Focus: Elementary mathematics and beyond
Re: 1st and 2nd Derivatives ..

If it's any help:





greg1313 is offline  
May 13th, 2010, 09:50 PM   #9
Senior Member
 
Joined: Mar 2010
From: Melbourne

Posts: 178
Thanks: 0

Re: 1st and 2nd Derivatives ..

Hm,

Still cant see how it works.

From my answer:

(e^x)(x^-1)+(e^x)(-1x^-2)-(e^x)(-1x^-2)+(e^x)(2x^-3)

Broken into two:

(e^x)(x^-1)+(e^x)(-1x^-2) (i)

'-'

(e^x)(-1x^-2)+(e^x)(2x^-3) (ii)

(i)

(e^x)(x^-1) - (e^x)(x^-2)

(ii)

-(e^x)(x^-2) + (2e^x)(x^-3)

--

Rejoining them:

(e^x)(x^-1)-(e^x)(x^-2)--(e^x)(x^-2)+(2e^x)(x^-3)

(e^x)(x^-1)-(e^x)(x^-2)+(e^x)(x^-2)+(2e^x)(x^-3)

--

Which does not work out ? Any idea where I'm going wrong ?

wulfgarpro.
wulfgarpro is offline  
May 13th, 2010, 10:22 PM   #10
Global Moderator
 
greg1313's Avatar
 
Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,958
Thanks: 1146

Math Focus: Elementary mathematics and beyond
Re: 1st and 2nd Derivatives ..

Start with:
(e^x)(x^-1)

Differentiate:
(e^x)(x^-1)-(e^x)(x^-2)

-- differentiate again:
(e^x)(x^-1)-(e^x)(x^-2) - ((e^x)(x^-2) - (e^x)(2x^-3))

Watch the signs!
greg1313 is offline  
Reply

  My Math Forum > High School Math Forum > Algebra

Tags
1st, 2nd, derivatives



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Are partial derivatives the same as implicit derivatives? johngalt47 Calculus 2 November 23rd, 2013 11:44 AM
Please Help.. Derivatives dV/dR - dV/dt fantom2012 Calculus 3 May 6th, 2012 07:53 PM
Derivatives-Help Arley Calculus 5 April 13th, 2012 06:56 PM
derivatives help r-soy Calculus 2 November 27th, 2010 08:26 AM
derivatives Calavan11 Calculus 2 December 11th, 2009 04:42 AM





Copyright © 2019 My Math Forum. All rights reserved.