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ladyrae
Jun12-04, 01:52 PM
Is this right? I'm not sure if I'm on the right track.

Using the definition of derivate find f ` (x)

f (x) = x - (a/x) , where a is a constant

f ` (x) = lim h->0 [(f(x+h)) – (f(x))]/h

= lim h->0 [((x+h) –(a/(x+h)) -(x-(a/x))]/h
= lim h->0 ([((x+h) –(a/(x+h)) -(x-(a/x))]/h) . (x(x+h))/(x(x+h))
= lim h->0 ((x^3)+(h^2x)-(ax)-(x^3)-(x^2h)+(ax)+(ah))/(h(x^2+xh))
= lim h->0 (h(hx-x^2+a))/(h(x^2+xh)) = (-x^2+a)/x^2

jdavel
Jun12-04, 02:07 PM
ladyrae,

Well, you don't seem to be getting the right answer, so you must not be on the right track. A few suggestions:

The notation you're showing us is almost impossible to read; I hope what you're looking at is a little more legible.

Try splitting this up by using the definition to that the derivative of the sum of two functions equals the sum of the derivatives of each function. Then you can do x and -a/x separately and just put them together when you're done.

Then show that the derivative of a constant times a function equals the constant times the derivative of the function. Then you can do 1/x and just multiply by -a when you're done.

arildno
Jun12-04, 02:08 PM
= lim h->0 ((x^3)+(h^2x)-(ax)-(x^3)-(x^2h)+(ax)+(ah))/(h(x^2+xh))
while correct, this move does not really help you; it complicates your expression!

= lim h->0 ((x^3)+(h^2x)-(ax)-(x^3)-(x^2h)+(ax)+(ah))/(h(x^2+xh))

Here, you've been led astray by your complicated expression!
You should have:
= lim h->0 ((x^3)+2hx^2+(h^2x)-(ax)-(x^3)-(x^2h)+(ax)+(ah))/(h(x^2+xh))

This gives f'(x)=(a+x^2)/x^2