Trouble with Derivatives

**Jay23456** · May 9th 2010, 06:55 PM

I've learned the short cut to obtain a derivate (nX^n-1) but some questions on exams ask to find the answer using the definition

Lim f(x+h)-f(x)
h->0 h

now I am terribly lost how to solve this is say f(x) = sqrt (x+2)

what I would think to do is rewrite it was

(x+2)^1/2 + h) - (x+2)^1/2
h

this is where I go astray...
I am totally baffled on what the next step would be...if it was x^2 I would simply factor it out and then divide by h but the 1/2 is totally making my thought process come to a dead halt. Please tell me if my approach is wrong or what the next step would be. Thanks

**Prove It** · May 9th 2010, 07:00 PM

Originally Posted by Jay23456

I've learned the short cut to obtain a derivate (nX^n-1) but some questions on exams ask to find the answer using the definition

Lim f(x+h)-f(x)
h->0 h

now I am terribly lost how to solve this is say f(x) = sqrt (x+2)

what I would think to do is rewrite it was

(x+2)^1/2 + h) - (x+2)^1/2
h

this is where I go astray...
I am totally baffled on what the next step would be...if it was x^2 I would simply factor it out and then divide by h but the 1/2 is totally making my thought process come to a dead halt. Please tell me if my approach is wrong or what the next step would be. Thanks

$f(x) = \sqrt{x + 2}$

$f(x + h) = \sqrt{x + h + 2}$ .

$\frac{f(x + h) - f(x)}{h} = \frac{\sqrt{x + h + 2} - \sqrt{x + 2}}{h}$

$= \left(\frac{\sqrt{x + h + 2} - \sqrt{x + 2}}{h}\right)\left( \frac{\sqrt{x + h + 2}+\sqrt{x + 2}}{\sqrt{x + h + 2} + \sqrt{x + 2}}\right)$

$= \frac{(x + h + 2) - (x + 2)}{h(\sqrt{x + h + 2} + \sqrt{x + 2})}$

$= \frac{h}{h(\sqrt{x + h + 2} + \sqrt{x + 2})}$

$= \frac{1}{\sqrt{x + h + 2} + \sqrt{x + 2}}$ .

So $f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$

$= \lim_{h \to 0}\frac{1}{\sqrt{x + h + 2} + \sqrt{x + 2}}$

$= \frac{1}{\sqrt{x + 2} + \sqrt{x + 2}}$

$= \frac{1}{2\sqrt{x + 2}}$ .

**Jay23456** · May 9th 2010, 07:05 PM

so when there a square root in the numerator you multiply by the conjugate?

**Prove It** · May 9th 2010, 07:06 PM

Originally Posted by Jay23456

so when there a square root in the numerator you multiply by the conjugate?

Multiply top and bottom by the top's conjugate, yes.

**Jay23456** · May 9th 2010, 07:22 PM

Okay thanks and what would you do if say f(x)= 5/(2-x) ?

**HallsofIvy** · May 10th 2010, 02:53 AM

$\lim_{h\to 0}\frac{\frac{5}{2-(x+h)}-\frac{5}{2-x}}{h}$

First simplify $\frac{5}{2-(x+h)}- \frac{5}{2- x}$ by subtracting the fractions- get common denominator by multiplying numerator and denominator of the first fraction by 2- x and the second fraction by 2-x-h:
$\frac{5(2-x)}{(2-x)(2-x-h)}- \frac{5(2-x-h)}{(2-x)(2-x-h)}$ $= \frac{10- 5x- 10+ 5x- 5h}{(2- x)(2- x- h)}$ .

Can you finish it?

**Jay23456** · May 10th 2010, 06:39 AM

Where did -5h come from? is -5 times -h = +5h?

**Prove It** · May 10th 2010, 06:40 AM

Originally Posted by Jay23456

Where did -5h come from? is -5 times -h = +5h?

Yes, it should be $+5h$ .

**Jay23456** · May 10th 2010, 06:48 AM

So I got $5h/(2-x)(2-x-h)$

is that correct? also i'm not sure if the above common denominator is correct shouldn't the denominator be multiplied by (2-x+h)?

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