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Why? Question about Differentiation Exersice.

  • Thread starter alba_ei
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  • #1
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the exersice is from stewart's book Ch 5.4 #3
i have the function [tex] y = \frac{x}{a^2 \sqrt(a^2-x^2)} [/tex]
this can be simplified like this

[tex] y = c*R(x) [/tex]

where R(x) = P(x)/Q(x) my question is

in the step of -P(x)*Q'(x) why is not zero????

if the formula says [tex] U^n = nU^n ^(-1) du [/tex]

in the exaple would be [tex] U^n = nU^n ^(-1) du [/tex]
(everything)*(2a*0+2x*1) so that zero makes all zero!!

so our answer is [tex] y' = \frac{1}{a^2 \sqrt(a^2-x^2)} [/tex], doesnt it?

i saw the real answer and its [tex] y' = \frac{1}{\sqrt((a^2-x^2))^3} [/tex]

im i wrong in the formula that i use for [tex] U^n [/tex]
 
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Answers and Replies

  • #2
cristo
Staff Emeritus
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i have the function [tex] y = \frac{x}{a^2 \sqrt(a^2-x^2)} [/tex]
this can be simplified like this

[tex] y = c*R(x) [/tex]

where R(x) = P(x)/Q(x) my question is

in the step of -P(x)*Q'(x) why is not zero????

if the formula says [tex] U^n = nU^n ^-1 du [/tex]

in the exaple would be [tex] U^n = nU^n ^-1 du [/tex]
(everything)*(2a*0+2x*1) so that zero makes all zero!!

so our answer is [tex] y' = \frac{1}{a^2 \sqrt(a^2-x^2)} [/tex], doesnt it?

i saw the real answer and its [tex] y' = \frac{1}{\sqrt((a^2-x^2))^3} [/tex]

im i wrong in the formula that i use for [tex] U^n [/tex]
Your denominator is [tex]a^2\sqrt{a^2-x^2}=a^2(a^2-x^2)^{1/2}[/tex]. Here, a is a constant, and you have a function of a function of x, so you will need to use the chain rule. Your function [itex]U=a^2-x^2[/itex] and n=1/2. Using the chain rule on this you will have [tex]\frac{dU}{dx}=\frac{dU}{d(x^2)}\frac{d}{dx}(x^2)[/tex]
 
  • #3
Gib Z
Homework Helper
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And what formula says [tex]u^n = nu^{n-1} du [/tex]? Im not sure that makes sense. [tex]y=u^n[/tex] Then [tex]dy/du = nu^{n-1}[/tex]
 
  • #4
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And what formula says [tex]u^n = nu^{n-1} du [/tex]? Im not sure that makes sense. [tex]y=u^n[/tex] Then [tex]dy/du = nu^{n-1}[/tex]
well you understand what i tried to say so why dont you help me
 
  • #5
HallsofIvy
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He was trying to help you. Perhaps not with the that particular problem but with something far more important. Many students have an unfortunate tendency to write "= " when they mean "this is the result" of some operation. Gib Z was pointing out how important it is in mathematics to be careful and precise. It is simply not true that [itex]u^n= nu{n-1}du[/itex] and no book ever told you that. And the problem with expecting people to "understand what you tried to say" is that we don't know whether you just wrote it carelessly or honestly don't understand the difference yourself.

That problem is magnified when you write
my question is
in the step of -P(x)*Q'(x) why is not zero????
if the formula says
[tex] U^n = nU^{n-1} du [/tex]
in the exaple would be [tex] U^n = nU^{n-1} [/tex]
(everything)*(2a*0+2x*1) so that zero makes all zero!!
I can't make any sense of that at all! Why is what "not zero". Where did you get (everything)*(2a*0+ 2x*1)? If you are referring to the 0 in "2a*0" then obviously (2a*0+ 2x*1)= 2x- it doesn't make "all" 0.

Since you say
in the step of -P(x)*Q'(x)
, I take it you are talking about the derivative of the denominator [itex]a^2\sqrt{a^2- x^2}= a^2(a^2- x^2)^{\frac{1}{2}}[/itex]. The [itex]a^2[/itex] is a constant so we don't need to differentiate that. The derivative of [itex](a^2- x^2)^{\frac{1}{2}}[/itex] is [itex]\frac{1}{2}(a^2- x^2)^{-\frac{1}{2}}[/itex] times the derivative of [itex]a^2- x^2[/itex] which is -2x. The derivative of the denominator is
[tex]-2a^2x(a^2-x^2)^{-\frac{1}{2}}[/tex].
Is that what you got?
 
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