testing header
Math Goodies is a free math help portal for students, teachers, and parents.
Free Math
Newsletter
 
 
Interactive Math Goodies Software

Buy Math Goodies Software
testing left nav
Math Forums @ Math Goodies
Math Forums @ Math Goodies
Home | Profile | Active Topics | Members | Search | FAQ
Username:
Password:
Save Password
Forgot your Password?

 All Forums
 Homework Help Forums
 Pre-Calculus and Calculus
 power/chain rules
 New Topic  Topic Locked
 Printer Friendly
Author Previous Topic Topic Next Topic  

uberclay
Advanced Member

Canada
159 Posts

Posted - 10/13/2007 :  01:50:34  Show Profile
I just finished a lesson on tangents which was capped off with the Power Rule and the Chain Rule for derivatives.

I cannot figure out how to differentiate the following, given the rules of this lesson. Can someone please point in the right direction.

I'm really unsure about this one. In all of the practice questions I was able to eliminate the denominator.

Q: differentiate y = (8x^4 - 5x^2 - 2) / (4x^3)

A:
y = (8x^4 / 4x^3) - (5x^2 / 4x^3) - (2 / 4x^3)
y = 2x - (5/4)x^-1 - (2/4)x^-3

y' = 2 + (5/4)x^-2 + (3/2)x^-4

Q: differentiate y = (5x)^(1/2) - (x/5)^(1/2)

A:
Similar problem with this one. None of the practice questions had coefficients in the radicand.

I've split this one into two parts, (5x)^(1/2) and (x/5)^(1/2).

for s = (5x)^(1/2)

let t = 5x, so that s = t^(1/2)

dt/dx = 5
ds/dt = (1/2)t^(-1/2)

chain rule: ds/dx = dt/dx * ds/dt, t = 5x
therefore
ds/dx = 5 * (1/2)(5x)^(-1/2)
ds/dx = 5 / 2(5x)^(1/2)

for u = (x/5)^(1/2)

let v = x/5, so that u = v^(1/2)

dv/dx = 1/5
du/dv = (1/2)v^(-1/2)

chain rule: du/dx = dv/dx * du/dv, u = (x/5)^(1/2)
therefore
du/dx = (1/5) * (1/2)[(x/5)^(1/2)](-1/2)
du/dx = 5^(1/2) / 10x^(1/2)

TF: dy/dx = [5 / 2(5x)^(1/2)] - [5^(1/2) / 10x^(1/2)]

I am very sure that both these answers are way off base.

Thanks for your time.

Edited by - uberclay on 10/14/2007 01:28:11
Go to Top of Page

skeeter
Advanced Member

USA
5634 Posts

Posted - 10/13/2007 :  07:32:50  Show Profile
first one is fine ... no need for the chain rule (why use it when you don't have to?)

second one is fine, although you took a long road to arrive your derivative.

y = (5x)1/2 - (x/5)1/2

dy/dx = (1/2)(5x)-1/2*5 - (1/2)(x/5)-1/2*(1/5)

dy/dx = 5/[2(5x)1/2] - 51/2/(10x1/2)

which can be simplified further ...

dy/dx = (2/5)(5/x)
Go to Top of Page

uberclay
Advanced Member

Canada
159 Posts

Posted - 10/15/2007 :  23:13:43  Show Profile
Thank you for checking those for me Skeeter.

My instructor expects me to take the long way, at times, to show a complete understanding of the concepts in the lesson. I milled over that second question tonight and did it w/o the Chain Rule, using that common factor David exposed. This time I got a different looking derivative but soon found that they were equal.

David, I see what you are saying about joining at the finish in question 1. I'll take care of that tomorrow.

g'night.
Go to Top of Page

uberclay
Advanced Member

Canada
159 Posts

Posted - 10/16/2007 :  15:32:11  Show Profile
Is

dy/dx = (8x^4 + 5x^2 + 6) / 4x^4

a suitable answer for question 1?

Thanks
Go to Top of Page

skeeter
Advanced Member

USA
5634 Posts

Posted - 10/16/2007 :  18:42:44  Show Profile
yes
Go to Top of Page
  Previous Topic Topic Next Topic  
 New Topic  Topic Locked
 Printer Friendly
Jump To:
Math Forums @ Math Goodies © 2000-2004 Snitz Communications Go To Top Of Page
This page was generated in 0.04 seconds. Snitz Forums 2000
testing footer
About Us | Contact Us | Advertise with Us | Facebook | Blog | Recommend This Page




Copyright © 1998-2014 Mrs. Glosser's Math Goodies. All Rights Reserved.

A Hotchalk/Glam Partner Site - Last Modified 21 Nov 2014