Trouble with Derivatives

Apr 2010
29
0
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

MHF Helper
Aug 2008
12,883
4,999
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
\(\displaystyle f(x) = \sqrt{x + 2}\)

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



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

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

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

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

\(\displaystyle = \frac{1}{\sqrt{x + h + 2} + \sqrt{x + 2}}\).



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

\(\displaystyle = \lim_{h \to 0}\frac{1}{\sqrt{x + h + 2} + \sqrt{x + 2}}\)

\(\displaystyle = \frac{1}{\sqrt{x + 2} + \sqrt{x + 2}}\)

\(\displaystyle = \frac{1}{2\sqrt{x + 2}}\).
 
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Reactions: Bruno J.
Apr 2010
29
0
so when there a square root in the numerator you multiply by the conjugate?
 
Apr 2010
29
0
Okay thanks and what would you do if say f(x)= 5/(2-x) ?
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
\(\displaystyle \lim_{h\to 0}\frac{\frac{5}{2-(x+h)}-\frac{5}{2-x}}{h}\)

First simplify \(\displaystyle \frac{5}{2-(x+h)}- \frac{5}{2- x}\) 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:
\(\displaystyle \frac{5(2-x)}{(2-x)(2-x-h)}- \frac{5(2-x-h)}{(2-x)(2-x-h)}\)\(\displaystyle = \frac{10- 5x- 10+ 5x- 5h}{(2- x)(2- x- h)}\).

Can you finish it?
 
Apr 2010
29
0
Where did -5h come from? is -5 times -h = +5h?
 
Apr 2010
29
0
So I got \(\displaystyle 5h/(2-x)(2-x-h)\)

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)?