My Math Forum  

Go Back   My Math Forum > College Math Forum > Differential Equations

Differential Equations Ordinary and Partial Differential Equations Math Forum


Thanks Tree1Thanks
  • 1 Post By fysmat
Reply
 
LinkBack Thread Tools Display Modes
September 1st, 2016, 05:23 AM   #1
Newbie
 
Joined: Sep 2016
From: York

Posts: 2
Thanks: 0

Quick differentiation problem

Hi, I'm Josh, just had a problem I've been thinking about and I think I'm missing a piece of... so...

It's a bit of implicit differentiation, I just need to know how to implicitly differentiate half of an equation. Which is sqrt(x+y), differentiating with respect to x. obviously ( sqrt of x ) prime is 1/2(x)^-.5, I just was confused about the chain rule-- does it come into play at all here? With the y? And so...
thanks.
AguywhoneedsHelp is offline  
 
September 1st, 2016, 07:12 AM   #2
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: USA

Posts: 2,529
Thanks: 1389

$\dfrac {d}{dx} (x+y) = 1$

$\dfrac {d}{dy} (x+y) = 1$

so in both cases the factor caused by the application of the chain rule is simply 1.
romsek is offline  
September 1st, 2016, 11:26 AM   #3
Newbie
 
Joined: Sep 2016
From: York

Posts: 2
Thanks: 0

I want to know why though. Like what happens to the dy? Is it just not there at all?
AguywhoneedsHelp is offline  
September 1st, 2016, 11:33 AM   #4
Senior Member
 
fysmat's Avatar
 
Joined: Dec 2013
From: some subspace

Posts: 212
Thanks: 72

Math Focus: real analysis, vector analysis, numerical analysis, discrete mathematics
If we have $\displaystyle y = y(x)$ and we have some function, say $\displaystyle x^2 + y^2 = 9$ (a circle), then the implicit derivative of this with respect to $\displaystyle x$ is:

$\displaystyle 2x + 2yy' = 0$.

$\displaystyle 2x$ is the derivative of $\displaystyle x^2$
$\displaystyle 2yy'$ is the derivative of $\displaystyle y^2 = y(x)^2$ by using the chain rule.
$\displaystyle 0$ is the derivative of $\displaystyle 9$.

A warning, though, implicit differentiation can be dangerous if you don't know whether or not the $\displaystyle y$ really is a function.
Thanks from topsquark
fysmat is offline  
Reply

  My Math Forum > College Math Forum > Differential Equations

Tags
differentiation, problem, quick



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Quick problem...Need help MathMathHelloMath Algebra 3 January 7th, 2011 11:42 PM
Quick problem qynn Elementary Math 2 December 6th, 2010 03:22 PM
Quick canceling problem Hobstweedl Algebra 1 September 14th, 2007 06:00 AM
Quick differentiation question gruffyddd Calculus 3 September 11th, 2007 06:47 AM
Quick problem...Need help MathMathHelloMath Calculus 3 December 31st, 1969 04:00 PM





Copyright © 2019 My Math Forum. All rights reserved.