# Trouble with Derivatives

#### 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

#### Prove It

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

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

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

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

Bruno J.

#### Jay23456

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

#### Prove It

MHF Helper
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

#### Jay23456

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

#### HallsofIvy

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

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?

#### Jay23456

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

#### Prove It

MHF Helper
Where did -5h come from? is -5 times -h = +5h?
Yes, it should be $$\displaystyle +5h$$.

#### Jay23456

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)?