# Thread: find derivative of sqrtx...

1. ## find derivative of sqrtx...

Use the definition to find the derivative of f(x)= $\sqrt{x}$, for >0. Is f differential at zero? Explain.

I believe the def we need is let f: D->R with $x_0$ an accumulation point of D and $x_0$ $\in$ D.with x not equal to x_0, define T(x)=f(x)-f(x_0)/x-x_0. The function f is said to be differentiable at x_0 iff T has a limit at x_0 and we write lim_x->x_0 T(x)=f'(x_0). ..I understand it I'm just sure how to show a proof and stuff..

2. ## Derivative

Hello,

If you would like to find the derivative of the function $f(x)= \sqrt{x}$, then perhaps you could follow the 3-step rule:

Let $f$ be a function and $x_{0}$ a point in the domain, then do as follows:

1) Compute $\triangle y=f(x_{0}+h)-f(x_{0})$

2) Compute and reduce $\frac{\triangle y}{h}=\frac{f(x_{0}+h)-f(x_{0})}{h}$

3) Let $h\to \infty$ in the reduced result from 2). If this limit exists then this limit will be the derivative, and the function $f$ differentiable.

3. Originally Posted by surjective
Hello,

If you would like to find the derivative of the function $f(x)= \sqrt{x}$, then perhaps you could follow the 3-step rule:

Let $f$ be a function and $x_{0}$ a point in the domain, then do as follows:

1) Compute $\triangle y=f(x_{0}+h)-f(x_{0})$

2) Compute and reduce $\frac{\triangle y}{h}=\frac{f(x_{0}+h)-f(x_{0})}{h}$

3) Let $h\to \infty$ in the reduced result from 2). If this limit exists then this limit will be the derivative, and the function $f$ differentiable.
isnt that basically the same def i posted?

4. Originally Posted by surjective
Hello,

If you would like to find the derivative of the function $f(x)= \sqrt{x}$, then perhaps you could follow the 3-step rule:

Let $f$ be a function and $x_{0}$ a point in the domain, then do as follows:

1) Compute $\triangle y=f(x_{0}+h)-f(x_{0})$

2) Compute and reduce $\frac{\triangle y}{h}=\frac{f(x_{0}+h)-f(x_{0})}{h}$

3) Let $h\to \infty$ in the reduced result from 2). If this limit exists then this limit will be the derivative, and the function $f$ differentiable.
would that give me $\sqrt{h}$/h ?

5. Originally Posted by alice8675309
would that give me $\sqrt{h}$/h ?
Firstly, yes your definition is correct. You need to let $h \to 0$ instead of $\infty$, though..

Your result is not correct - $\sqrt{x+h} - \sqrt{x} \neq \sqrt{x+h-x}$... Instead, multiply by conjugate:

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

And from there, simplify and then let $h \to 0$ to get the result.

6. Originally Posted by Defunkt
Firstly, yes your definition is correct. You need to let $h \to 0$ instead of $\infty$, though..

Your result is not correct - $\sqrt{x+h} - \sqrt{x} \neq \sqrt{x+h-x}$... Instead, multiply by conjugate:

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

And from there, simplify and then let $h \to 0$ to get the result.
ok so now i have lim_h->0 1/ $\sqrt{x+h}$+ $\sqrt{x}$ and then as the limit of h goes to 0 we have 1/2 $\sqrt{x}$, which means yes f is differentiable at zero, b/c as h goes to zero the derivative equals 1/2 $\sqrt{x}$

7. Also, to put a fraction in math tags:

\frac{\sqrt{x+h} - \sqrt{x}}{h} produces $\frac{\sqrt{x+h} - \sqrt{x}}{h}$