Using definition to find derivative

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December 4th 2009, 10:59 AM
thekrown

Using definition to find derivative

5. c) If g(x) = 1 / square root (2x+1), use the definition of the derivative to find g prime (1).

I would greatly appreciate a short walk through on this number. I've divided the steps into 5.

Step 1 was to arrange everything so it looks like the formula

g(x+h) - g prime(x) / h

Step 2 was to put everything on common denominator.

Step 3 rationalize - this is where I get stuck

The teacher went over it but just "said the nominator becomes -2h". I tried cracking this and cannot get -2h. There must be something wrong with my algebra, I've tried many many times and by substituing x=1 I still can't.

Step 4 Is to simplify and we should end up with -1 / 3^(3/2).
December 4th 2009, 03:20 PM
skeeter

$g'(x) = \lim_{x \to 0} \frac{1}{h} \left(\frac{1}{\sqrt{2(x+h)+1}} - \frac{1}{\sqrt{2x+1}}\right)$

$g'(x) = \lim_{x \to 0} \frac{1}{h} \left(\frac{\sqrt{2x+1} - \sqrt{2(x+h)+1}}{\sqrt{2(x+h)+1} \cdot \sqrt{2x+1}}\right)$

$g'(x) = \lim_{x \to 0} \frac{1}{h} \left(\frac{\sqrt{2x+1} - \sqrt{2(x+h)+1}}{\sqrt{2(x+h)+1} \cdot \sqrt{2x+1}}\right) \cdot \frac{\sqrt{2x+1} + \sqrt{2(x+h)+1}}{\sqrt{2x+1} + \sqrt{2(x+h)+1}}$

$g'(x) = \lim_{x \to 0} \frac{1}{h} \left(\frac{(2x+1) - [2(x+h)+1]}{\sqrt{2(x+h)+1} \cdot \sqrt{2x+1}\left[\sqrt{2x+1} + \sqrt{2(x+h)+1}\right]}\right)$

$g'(x) = \lim_{x \to 0} \frac{1}{h} \left(\frac{-2h}{\sqrt{2(x+h)+1} \cdot \sqrt{2x+1}\left[\sqrt{2x+1} + \sqrt{2(x+h)+1}\right]}\right)$

$g'(x) = \lim_{x \to 0} \frac{-2}{\sqrt{2(x+h)+1} \cdot \sqrt{2x+1}\left[\sqrt{2x+1} + \sqrt{2(x+h)+1}\right]}$

$g'(x) = \frac{-2}{(2x+1)(2\sqrt{2x+1})}$

$g'(x) = -\frac{1}{(2x+1)^{\frac{3}{2}}}$
December 4th 2009, 04:03 PM
thekrown

I am able to get -2h and of course the h cancels on steps 5 to 6.

Now the transition from step 6 to 7 is odd. I notice we multiply outer with outer and inner with inner to obtain what you have on the step 7 denominator, but how come there is no sign of the inner*outer and outer*inner multiplications?
December 4th 2009, 04:06 PM
skeeter

Quote:

Originally Posted by thekrown

I am able to get -2h and of course the h cancels on steps 5 to 6.

Now the transition from step 6 to 7 is odd. I notice we multiply outer with outer and inner with inner to obtain what you have on the step 7 denominator, but how come there is no sign of the inner*outer and outer*inner multiplications?

because they sum to 0.

$(a - b)(a + b) = a^2 + \textcolor{red}{ab - ab} + b^2$
December 4th 2009, 04:22 PM
thekrown

Okay I understand how we get -2 / (2x+1) (2 square root 2x+1) but I can't get the final answer you get.

I know it has something to do with the second term in the denominator but I can't simplify it into the first term to obtain an exponent of 3/2.

I've tried substituting for one (g prime (1)). and get:

Step 1) -2 / square root 3 * square root 3 ( square root 3 + square root 3).

Step 2) -2 / 3 ( square root 3 + square root 3)

This is where I'm stuck.
December 4th 2009, 04:29 PM
skeeter

Quote:

Originally Posted by thekrown

Okay I understand how we get -2 / (2x+1) (2 square root 2x+1) but I can't get the final answer you get.

I know it has something to do with the second term in the denominator but I can't simplify it into the first term to obtain an exponent of 3/2.

I've tried substituting for one (g prime (1)). and get:

Step 1) -2 / square root 3 * square root 3 ( square root 3 + square root 3).

Step 2) -2 / 3 ( square root 3 + square root 3)

This is where I'm stuck.

$-\frac{2}{3(\sqrt{3} + \sqrt{3})}$

note that $\sqrt{3} + \sqrt{3} = 2\sqrt{3}$ ...

$-\frac{2}{3(2\sqrt{3})} = -\frac{1}{3\sqrt{3}}$

also, note that $3\sqrt{3} = 3^1 \cdot 3^{\frac{1}{2}} = 3^{\frac{3}{2}}$
December 4th 2009, 04:32 PM
thekrown

Yes thank you, I did not see that. I was able to solve it by altering (square root 3 + square root 3) into (3+3)^1/2 and then (3x2)^1/2.

My algebra is incredibly rusty but if what I did is correct then the 2's cancel and we get

-1 / 3 * 3^1/2 = -1 / 3^3/2

Whew!

I do prefer your method. Thanks for the help!