# Thread: derivative of a function problem

1. ## derivative of a function problem

i'm having a major brain clot,

the derivative of this function

f(x)= √x + 1

using the limit defintion

thank you so much!

2. You don't have to use limits to find this derivative.

Recall that:

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

So:

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

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

3. Originally Posted by sondre
i'm having a major brain clot,

the derivative of this function

f(x)= √x + 1

using the limit defintion

thank you so much!
$\displaystyle f(x) = \sqrt{x}+1$

derivative:
$\displaystyle =\lim_{h\to 0} \frac{\sqrt{x+h}+1-\sqrt{x}-1}h$

$\displaystyle =\lim_{h\to 0} \frac{\sqrt{x+h}-\sqrt{x}}h * \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}$

$\displaystyle =\lim_{h\to 0} \frac{h+x-x}{h(\sqrt{x+h}+\sqrt{x})}$

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

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

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

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

4. Hello, sondre!

I assumed that the square root is over the entire $\displaystyle x + 1$

Differentate: .$\displaystyle f(x)\:= \:\sqrt{x + 1}$

using the limit defintion: .$\displaystyle f'(x) \;=\;\lim_{h\to0}\frac{f(x+h) - f(x)}{h}$
There are four steps . . .

(1) Find $\displaystyle f(x+h)\!:\quad f(x+h) \;=\;\sqrt{(x+h)+1}$

(2) Subtract $\displaystyle f(x)\!:\quad f(x+h)-f(x) \;=\;\sqrt{x+h+1} - \sqrt{x+1}$

. . Multiply top and bottom by $\displaystyle \sqrt{x+h+1} + \sqrt{x+1}$

. . . . $\displaystyle \frac{\sqrt{x+h+1} - \sqrt{x+1}}{1}\cdot\frac{\sqrt{x+h+1} + \sqrt{x+1}}{\sqrt{x+h+1} + \sqrt{x+1}}$

. . . . . $\displaystyle =\;\frac{(x+h+1) - (x+1)}{\sqrt{x+h+1} + \sqrt{x+1}} \;=\;\frac{h}{\sqrt{x+h+1}-\sqrt{x+1}}$

(3) Divide by $\displaystyle h\!:\quad \frac{1}{h}\cdot\frac{h}{\sqrt{x+h+1} + \sqrt{x+1}} \;=\;\frac{1}{\sqrt{x+h+1} + \sqrt{x+1}}$

(4) Take limit: . $\displaystyle \lim_{h\to0}\left[\frac{1}{\sqrt{x +h+1} + \sqrt{x+1}}\right] \;=\;\frac{1}{\sqrt{x+1} + \sqrt{x+1}}$

Therefore: .$\displaystyle \boxed{f'(x) \;=\;\frac{1}{2\sqrt{x+1}}}$

5. Originally Posted by Soroban
Hello, sondre!

I assumed that the square root is over the entire $\displaystyle x + 1$
Oh, thats a good call, I hadn't thought of that >.<