# Differentiable

1. Sep 21, 2014

### Unusualskill

(a) State precisely the definition of: a function f is differentiable at a ∈ R.
(b) Prove that, if f is differentiable at a, then f is continuous at a. You may
assume that
f'(a) = lim {f(x)-f(a)/(x-a)} as x approaches a

(c) Assume that a function f is differentiable at each x ∈ R and also f(x) > 0
for all x ∈ R. Use the definition of the derivative and standard limit laws to
calculate the derivative of:
g(x) = (f(x))^0.25
in terms of f(x) and f'(x).

Im stuck at part(c)...Can guide me or show me how to start?thx alot!

2. Sep 21, 2014

### Fredrik

Staff Emeritus
You have to show us your work up to the point where you're stuck. It can't be right at the start, since the problem tells you how to start. (Use the definition of "derivative"). What have you tried to do to proceed from the point where you're stuck?

3. Sep 21, 2014

### Unusualskill

i do until

{lim (f(x))^0.25 - lim (f(a))^0.25 }/lim (x-a) as x approaches a

Anyway how to type my solutions here involving the limits( I mean like x approaches a can write it as x--->a)

4. Sep 21, 2014

### Fredrik

Staff Emeritus
The rule $\lim_{x\to a}\frac{F(x)}{G(a)}=\frac{\lim_{x\to a}F(x)}{\lim_{x\to a}G(x)}$ doesn't work so well when the right-hand side is 0/0.

I don't know what you're supposed to do here. The first thing that comes to mind is to do Taylor expansions. For example, $F(x)=F(a)+(x-a)F'(a)+O((x-a)^2)$ as $x\to a$. Do you think you would be allowed to do that here?

There's a FAQ post about LaTeX. You can also quote me to see how I'm doing it.

5. Sep 21, 2014

### HallsofIvy

The problem says "use the definition of the derivative and the standard limit laws" so you are clearly supposed to look at
$\lim_{h\to 0}\frac{f(x+h)^{0.25}- f(x)^{0.25}}{h}$. Rather than Taylor series I would use the fact that $x^4- y^4= (x- y)(x^3+ x^2y+ xy^2+ y^3)$, taking $x= f(x+ h)^{0.25}$ and $y= f(x)^{0.25}$. Then I would have
$$f(x+h)- f(x)= (f(x+h)^{0.25}- f(x)^{0.25})(f(x+h)^{0.75}+ f(x+h)^{0.50}f(x)^{0.25}+ f(x+ h)^{0.25}f(x)^{0.50}+ f(x)^{0.75}$$
which gives
$$f(x+h)^{0.25}- f(x)^{0.25}= \frac{f(x+h)- f(x)}{f(x+h)^{0.75}+ f(x+h)^{0.50}f(x)^{0.25}+ f(x+ h)^{0.25}f(x)^{0.50}+ f(x)^{0.75}}$$
and then
$$\frac{f(x+h)^{0.25}- f(x)^{0.25}}{h}= \frac{f(x+h)- f(x)}{h}\frac{1}{f(x+h)^{0.75}+ f(x+h)^{0.50}f(x)^{0.25}+ f(x+ h)^{0.25}f(x)^{0.50}+ f(x)^{0.75}}$$.

6. Sep 21, 2014

### Unusualskill

Thanks for your reply!But i am wondering is this the final answer?As the ques requires the ans in terms of f(x) and f '(x), how to get rid of the f(x+h) terms?

7. Sep 21, 2014

### Fredrik

Staff Emeritus
That's not the final answer. That formula is how you get rid of f(x+h). If we tell you the details, we would be giving you the complete solution.

8. Sep 21, 2014

### Unusualskill

Can gv me some hints or any theorem name for me to do research on?I have no clue at all.Thanks

9. Sep 21, 2014

### Fredrik

Staff Emeritus
What Halls told you is major clue. The rest is pretty straightforward. In what ways have you tried to use what he told you?

10. Sep 21, 2014

### Unusualskill

Oh! maybe I know already. After substituting what Halls told me into my limit, then i just break everything into pieces and apply limit law of sums?

11. Sep 21, 2014

### Fredrik

Staff Emeritus
Something like that, yes. (I don't know exactly what you did).

12. Sep 21, 2014

### Unusualskill

my answer f '(x)*(1/ (3(fx)^0.75 +(fx)^0.5) )

13. Sep 21, 2014

### Fredrik

Staff Emeritus
That looks correct.

When I said that Halls had given you a major clue, I was only looking at the part that was visible in the quote, before I clicked to expand. I thought that he had given you 75% of the solution, but he had in fact given you 95%. This is against the rules here, so you shouldn't expect to get this much information in your other threads. I think you need to read up on how the homework forum is supposed to work. It's not not OK to go "how do I get started" followed by "then what?" repeatedly until you have a complete answer. You are supposed to do the work.