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Differentiation from first principles

  • #1
Hey guys, I've got the following 2 mark question on a problem sheet, but I can't seem able to do it. I'd appreciate any help, thanks.

Differentiate, from first principles, the following:

[tex]y=\sqrt(a^2-x^2)[/tex]

I know I have to take the limit as δx tends to 0 of [(f(x+δx)- f(x)]/δx but can't seem to manipulate the resultant expression in any way to get closer to an answer. I have never used this Latex thingy before, so hopefully it comes out OK.

Thanks in advance!
 
Last edited:

Answers and Replies

  • #2
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Hey guys, I've got the following 2 mark question on a problem sheet, but I can't seem able to do it. I'd appreciate any help, thanks.

Differentiate, from first principles, the following:

[tex]y=\sqrt(a^2-b^2)[/tex]

I know I have to take the limit as δx tends to 0 of [(f(x+δx)- f(x)]/δx but can't seem to manipulate the resultant expression in any way to get closer to an answer. I have never used this Latex thingy before, so hopefully it comes out OK.
There is no x in the formula for your function. Are a and b constants or are they variables? If a and b are constants, then the graph of your function is a horizontal line, making the derivative zero.
 
  • #3
There is no x in the formula for your function. Are a and b constants or are they variables? If a and b are constants, then the graph of your function is a horizontal line, making the derivative zero.
Sorry, just a typo, [tex]b^2[/tex] should be [tex]x^2[/tex].

But a is just a constant yes.
 
  • #4
ehild
Homework Helper
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As for Latex thingies, use curly brackets{} for the argument of the square root.

[tex]y=\sqrt{a^2-x^2}[/tex]

It is a good method to eliminate difference of square roots by multiplying and dividing with the sum of the same square roots.
Write up the the expression (f(x+δx)-f(x))/δx with the square roots and you will see what I mean.


ehild
 
  • #5
Thank you very much for that, my final answer of

[tex]\frac{dy}{dx} =-\frac{x}{\sqrt{a^2-x^2}}[/tex]

Agrees with the answer obtained if you just differentiate 'normally'. Thanks again!
 

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