Differentiation from first principles

[Electrifying]

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:

y=\sqrt(a^2-x^2)

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!

 

[Mark44]

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:

y=\sqrt(a^2-b^2)

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.

 

[Electrifying]

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, b^2 should be x^2.

But a is just a constant yes.

 

[ehild]

As for Latex thingies, use curly brackets{} for the argument of the square root.

y=\sqrt{a^2-x^2}

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

 

[Electrifying]

Thank you very much for that, my final answer of

\frac{dy}{dx} =-\frac{x}{\sqrt{a^2-x^2}}

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