sqrt(x^2 - 9) / x

Question

I need to solve this using the definition of a derivative [f(a+h) - f(a)]/h

I can't seem to get it :\

Answer 1

The way to do this is to use a method like one used when working with complex numbers in the denominator.

(f(x+h) - f(x))/h = 1/h * [sqrt((x+h)^2-9)/(x+h) - sqrt(x^2-9)/x]

Put this all over a common denominator:

[x*sqrt((x+h)^2-9) - (x+h)*sqrt(x^2-9)]/hx(x+h)

Now, we want to use the fact that (u+v)(u-v) = u^2-v^2. This will help us eliminate all those pesky square roots. at least in the numerator. SO, multiply top and bottom by

x*sqrt((x+h)^2-9) + (x+h)*sqrt(x^2-9)

Cranking out the math, we now end up with the fraction

9h*(2x+h)/[hx(x+h) * (x*sqrt((x+h)^2-9) + (x+h)*sqrt(x^2-9)]

The lone h top and bottom cancels out as desired, leaving us with:

9*(2x+h)/[x(x+h) * (x*sqrt((x+h)^2-9) + (x+h)*sqrt(x^2-9)]

Now, we can take the limit as h goes to zero, with no pesky 0/0 problems. Just throw them away, giving us

9*(2x)/[x(x) * (x*sqrt((x)^2-9) + (x)*sqrt(x^2-9)]

= 18x/(x^2 * (2x * sqrt(x^2-9))

= 18x/(2x^3 * sqrt(x^2-9)

= 9 / (x^2 * sqrt(x^2-9)

*whew*

Answer 2

Thanks a lot!