Verified answer

For f(x) = 1 / √(x+2)

you get f(a) = 1 / √(a+2)

which is the same as (a+2)^(-1/2) where -1/2 is the power of (a+2)

Likewise f(a+h) = (a+2+h)^(-1/2)

Then, the difference quotient is lim as h-> 0 of { [f(a+h)-f(a) ] / h }

It is a bit easier to use f(a) = (a+2)^(-1/2) for this

and you get [f(a+h)-f(a) ] / h = [(a+2+h)^(-1/2) - (a+2)^(-1/2) ] /h

which is the same as [1/[(a+2+h)^(1/2) -1/(a+2)^(1/2)] / h

Add the fractions in the numerator of this complex fraction and you end up with

[(a+2)^(1/2) - (a+2+h)^(1/2)] / [(a+2+h)^(1/2)(a+2)^(1/2)] which is all over h

Then this is the same as

[(a+2)^(1/2) - (a+2+h)^(1/2)] /h [(a+2+h)^(1/2)(a+2)^(1/2)] --> since, in general, (1/x)/y = 1/xy

The crucial next step is to rationalize the numerator.

You do this by multiplying both your numerator and denominator by

[(a+2)^(1/2) + (a+2+h)^(1/2)]

This gives you [(a+2)^(1/2) - (a+2+h)^(1/2)] [(a+2)^(1/2) + (a+2+h)^(1/2)] (big numerator)

over h [(a+2+h)^(1/2)(a+2)^(1/2)] [(a+2)^(1/2) + (a+2+h)^(1/2)] (big denominator)

In your "big numerator" the expression simplifies to [(a+2)^(1/2)]^2 - [(a+2+h)^(1/2)] ^2

which in turn simplifies to (a+2)-(a+2+h) which equals a+2-a-2-h which is "- h"

Then,

you have -h / h [(a+2+h)^(1/2)(a+2)^(1/2)] [(a+2)^(1/2) + (a+2+h)^(1/2)] and the h's cancel out (!)

So you are trying to find

lim as h-> 0 of -1 / [(a+2+h)^(1/2)(a+2)^(1/2)] [(a+2)^(1/2) + (a+2+h)^(1/2)]

To simplify notation at this point let A = (a+2+h)^(1/2) and B = (a+2)^(1/2)

We will write lim as h-> 0 simply as "lim"

lim as h-> 0 of -1 / [(a+2+h)^(1/2)(a+2)^(1/2)] [(a+2)^(1/2) + (a+2+h)^(1/2)]

=- lim {-1/[(AB)(B+A)] } = lim (-1) / lim[(AB)(B+A)] =lim (-1 )/ [lim(AB)lim(B+A)] =

lim(-1) / (limAlimB)(limB+limA)

lim (-1 )= -1 (limit of a constant is that constant)

lim of A = B since lim A = lim h-> 0 of (a+2+h)^(1/2) which is (a+2) = B

lim of B = B because lim B = lim h-> 0 of (a+2)^(1/2) = (a+2)^(1/2) = B (h is not even involved)

then lim -1 / (limAlimB)(limB+limA) = -1 / BB(B+B) = -1 / BB(2B) = -1 / 2BBB = - 1/2B^3

But B = B = (a+2)^(1/2) so - 1/2B^3 = -1/2 [(a+2)^(1/2)]^3 = -1/2(a+2)^(3/2)

Answer: -1/2(a+2)^(3/2)

Note: I used the notation A and B to try to simplify the notation. When working this out with pencil and paper, it was simple to write the whole thing out, but not so simple to type it into this box...

Note 2: In the time it took me to type this thing in, 4 other answers were submitted...Wow--they must have some better software or something to get that much typed in as rapidly as they did...

Starting with what you have done:

{√(a + 2)/√(a + h + 2)*√(a + 2) - √(a + h + 2)/√(a + h + 2)*√(a + 2)}/h

We put the two numerators over the common denominator and include h as a factor in the denominator:

{√(a + 2) - √(a + h + 2)}/{h√(a + h + 2)*√(a + 2)}

Now multiply the above by 1 in form {√(a + 2) + √(a + h + 2)}/{√(a + 2) + √(a + h + 2)}:

{√(a + 2) - √(a + h + 2)}/{h√(a + h + 2)*√(a + 2)}{√(a + 2) + √(a + h + 2)}/{√(a + 2) + √(a + h + 2)}

I did the above step because it will make the numerator fit the form (u - v)(u + v). We know that his equals u² - v² and this will remove the squareroots from the numerator:

{(a + 2) - (a + h + 2)}/{h√(a + h + 2)*√(a + 2)[√(a + 2) + √(a + h + 2)]}

Now combine like terms in the numerator:

{-h}/{h√(a + h + 2)*√(a + 2)[√(a + 2) + √(a + h + 2)]}

Notice that -h/h becomes -1:

-1/{√(a + h + 2)*√(a + 2)[√(a + 2) + √(a + h + 2)]}

Now we may let h go to zero:

-1/{√(a + 2)*√(a + 2)[√(a + 2) + √(a + 2)]}

Use √(a + 2) + √(a + 2) = 2√(a + 2):

-1/{2√(a + 2)*√(a + 2)*√(a + 2)}

This equals your desired answer:

-1/{2(a + 2)^(3/2)}

Hello,

f(x) = 1/√(x + 2)

Since (α + β)(α - β)=α²-β², if α=√(a+2) and β=√(a+2+h):

[√(a + 2) - √(a + 2 + h)] × [√(a + 2) + √(a + 2 + h)] = (a + 2) - (a + 2 + h) = -h

Thus, we have the relation:

[√(a + 2) - √(a + 2 + h)] / h = -1 / [√(a + 2 + h) + √(a + 2)]

Let's calculate our limit now:

f'(a) = Lim (h→0) [f(a + h) - f(a)] / h

  = Lim (h→0) [1/√(a + 2 + h) - 1/√(a + 2)] / h

  = Lim (h→0) {√(a + 2) / √[(a + 2)(a + 2 + h)] - √(a + 2 + h) / √[(a + 2)(a + 2 + h)] } / h

  = Lim (h→0) {[√(a + 2) - √(a + 2 + h)] / h} / √[(a + 2)(a + 2 + h)]

By virtue of the identity above, we thus get:

f'(a) = Lim (h→0) -1 / {[√(a + 2 + h) + √(a + 2)] × √[(a + 2)(a + 2 + h)]}

  = -1 / {[√(a + 2) + √(a + 2)] × √[(a + 2)(a + 2)]}

  = -1 / [2√(a + 2) × (a + 2)]

  = -1 / [2(a + 2)^(3/2)]

QED

Methodically,

Dragon.Jade :-)

Actually, I think you're doing wonderfully! ...considering how problematic it is when

forming fractions by typing. I'm going to pick up your last line then modify it slightly

with different grouping symbols: [You've accurately defined your LCD already.]

= ( √(a+2) / √(a+h+2)*√(a+2) - √(a+h+2) / √(a+h+2)*√(a+2) ) / h =

( √(a+2) / LCD - √(a+h+2) / LCD ) / h =

[ {√(a+2) - √(a+h+2)} / LCD ] / h = ( √(a+2) - √(a+h+2) / [(h) LCD ]

I've done very little thus far except make things slightly easier to read.

Now RATIONALIZE THE NUMERATOR by mult top&bottom by the

conjugate √(a+2) + √(a+h+2) making the numerator just -h; you're home!

I'll use sqrt for √ . It is easier to type.

f(x) = 1/sqrt(x+2)

f(x+h)=1/sqrt(x+h+2)

f(x+h)-f(x) = 1/sqrt(x+h+2) - 1/sqrt(x+2)

f(x+h)-f(x) = [sqrt(x+2) - sqrt(x+h+2)] / sqrt(x+h+2)sqrt(x+2)

Multiply and divide by [sqrt(x+2)+ sqrt(x+h+2)]

f(x+h)-f(x) = [sqrt(x+2) - sqrt(x+h+2)] [sqrt(x+2)+ sqrt(x+h+2)] / sqrt(x+h+2)sqrt(x+2)[sqrt(x+2)+ sqrt(x+h+2)]

The numerator is (a-b)(a+b) = a^2-b^2

a= sqrt(x+2)

b=sqrt(x+h+2)

a^2 = x+2

b^2 = x+h+2

a^2-b^2 = x+2-x-h-2 = -h

f(x+h)-f(x) = -h / sqrt(x+h+2)sqrt(x)[sqrt(x+2)+ sqrt(x+h+2)]

[f(x+h)-f(x)]/h = -h / h sqrt(x+h+2)sqrt(x+2)[sqrt(x+2)+ sqrt(x+h+2)]

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

Let h approach 0

lim h-->0 f(x+h)-f(x)]/h = -1 / sqrt(x+2)sqrt(x+2) [ 2 sqrt(x+2)]

= -1/2(x+2)sqrt(x+2)

= -1/2(x+2)^(3/2)

f'(x) = -1/2(x+2)^(3/2)

f'(a) = -1/2(a+2)^(3/2)