My Math Forum Variables and unknown constants?

 Calculus Calculus Math Forum

 July 29th, 2011, 12:03 AM #1 Newbie   Joined: Dec 2010 Posts: 8 Thanks: 0 Variables and unknown constants? Hi, Through a maths course I have been learning about differentiating inverse trig functions. I just learnt the rule: $\begin{array}{c} y = {\sin ^{ - 1}}\left( {\frac{x}{a}} \right) \\ \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {{a^2} - {x^2}} }} \\ \end{array}$ I understand how to use the rule and how to prove the rule. I was recently shown a question $y= {\sin ^{ - 1}}\left( {\frac{x}{{{a^2}}}} \right)$ and to solve it they did $\begin{array}{c} \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {{{\left( {{a^2}} \right)}^2} - {x^2}} }} \\ = \frac{1}{{\sqrt {{a^4} - {x^2}} }} \\ \end{array}$ And then continued making it more elegant But what I don’t understand is why were they able to use the same rule and just sub it into the end? I mean if you got $y= {\sin ^{ - 1}}\left( {\frac{{{x^2}}}{a}} \right)$ You couldn’t just say that $\frac{{dy}}{{dx}}= \frac{1}{{\sqrt {{a^2} - {{\left( {{x^2}} \right)}^2}} }}$ Because that would be wrong because you need to use chain rule. So why can “a” be pretty much anything but “x” must remain as “x” for you to be able to use the rule? I tried to ask my teacher and he said it is because “a” is a constant. That confused me even more because I am not sure what the difference is between a variable and an unknown constant. Seeing as they both can take on any value. Do constants have some sort of property where you can replace them for anything in a rule and the rule still works while variables can’t be replaced without creating a new rule? Sorry if this is worded badly I am having a bit of trouble trying to explain why I am confused. I would really appreciate any help. Thank you
 July 29th, 2011, 12:08 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Variables and unknown constants? Since you are differentiating with respect to x, any expression or term not containing x, or a function of x, is treated as a constant during the differentiation.
July 29th, 2011, 12:21 AM   #3
Newbie

Joined: Dec 2010

Posts: 8
Thanks: 0

Re: Variables and unknown constants?

Quote:
 Originally Posted by MarkFL Since you are differentiating with respect to x, any expression or term not containing x, or a function of x, is treated as a constant during the differentiation.
Thanks for the quick reply. Ok so I understand that "a" is a constant but why is it that you can replace constants for anything (in this case ${a^2}$) and just sub the new value into $\frac{1}{{\sqrt {{a^2} - {x^2}} }}$ making it $\frac{1}{{\sqrt {{{\left( {{a^2}} \right)}^2} - {x^2}} }}$. But you can't do the same for variables and replace $x$ with ${x^2}$ and sub that into $\frac{1}{{\sqrt {{a^2} - {x^2}} }}$ making it $\frac{1}{{\sqrt {{a^2} - {{\left( {{x^2}} \right)}^2}} }}$. Is there some property of constants that allows you to do that?

 July 29th, 2011, 05:09 AM #4 Senior Member   Joined: Apr 2011 From: Recife, BR Posts: 352 Thanks: 0 Re: Variables and unknown constants? That's just like saying $\frac{d}{dx} (\frac{1}{x})= \frac{-1}{x^{2}}$ then $\frac{d}{dx} (\frac{1}{x^{2}})= \frac{-1}{x^{4}}$. You can only do such things with constants, because in your case $a^{2}$ is also a constant.
July 30th, 2011, 11:53 AM   #5
Global Moderator

Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,968
Thanks: 1152

Math Focus: Elementary mathematics and beyond
Re: Variables and unknown constants?

Quote:
 Originally Posted by delta_salt Is there some property of constants that allows you to do that?
As [color=#00AA00]MarkFL[/color] pointed out, the value of the constant remains the same no matter what the value of the variable we are differentiating with respect to (in this case, x). The constant's rate of change, with respect to x, is zero. Using a letter in place of a specific number allows us to generalize statements to accommodate any given constant, so rules of differentiation and the like can be written more compactly.

Any operation applied to a constant cannot involve x, otherwise the value of the constant would depend on x and we would have to account for that when we differentiate. Performing an operation on a constant that does not involve x results in another constant, i.e. its value does not depend on x, and its rate of change w.r.t.x. remains zero.

 July 31st, 2011, 02:33 AM #6 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Variables and unknown constants? Suppose the problem had been to differentiate $cos^{-1}\left(\frac{x}{5}\right)$, would you have had any problem using the fact that the derivative of $cos^{-1}\left(\frac{x}{a}\right)$ is $\frac{1}{\sqrt{a^2- x^2}}$ to say that the derivative of $cos^{-1}\left(\frac{x}{5}\right)$ is $\frac{1}{\sqrt{25- x^2}}$ just by replacing the symbol "a" by the number "5"? I would hope not- that is the whole point of algebra! We can use letters to represent numbers and replace them any way we wish. And if a represents "any number", the surely $a^2$ is also a number! We are simply replacing "a", which represents any number, by $a^2$ which also represents a number.
 July 31st, 2011, 06:01 AM #7 Newbie   Joined: Dec 2010 Posts: 8 Thanks: 0 Re: Variables and unknown constants? Thanks so much for the replies. I get it now!

 Tags constants, unknown, variables

,

### differentiation with unkown constants

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post 2902326 Calculus 2 March 3rd, 2014 06:52 PM ilgaar Number Theory 6 May 11th, 2013 09:42 AM Fruit Bowl Applied Math 2 January 31st, 2013 03:13 AM Yooklid Real Analysis 9 May 16th, 2010 02:38 PM E.L.Kim Calculus 3 August 18th, 2009 06:46 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top