Find the derivative from first principles

1. Jun 3, 2008

Asian Girl

1. The problem statement, all variables and given/known data
Find the derivative from first principles, i.e, from the definition of f ' as a limit, for f(x)= $$\sqrt{}a-bx, where a and b are positive constants. What are the domains of f and f ' ? 3. The attempt at a solution I solved: f(x) = \sqrt{a-bx} ---> f(x+h) = \sqrt{a-b(x+h)} lim h-->0 = \sqrt{a-b(x+h)} - \sqrt{a-bx} / h lim h-->0 = \sqrt{a-bx+bh-a+bx} / h lim h-->0 = \sqrt{bh} / h lim h-->0 = \sqrt{bh}* \sqrt{bh} / h\sqrt{bh} lim h-->0 = b/ \sqrt{bh} What did I do wrong here? Please help and thanks Last edited: Jun 3, 2008 2. Jun 3, 2008 Defennder Check this step. The former does not lead to the latter. You cannot "extend" the root from the left expression of the numerator to the right one. Instead, multiply the numerator (and the corresponding denominator) by its conjugate; ie. Conjugate of a+b is a-b and vice versa. 3. Jun 3, 2008 Asian Girl hmmm .... ok then i should do lim f(x) - g(x) = lim f(x) - lim g(x) = lim sf(a-b(x+h)) * sf(a-b(x+h)/ h sf(a-b(x+h)) * lim sf(a-bx) * sf(a-bx)/ h sf(a-bx) = lim a-bx+bh / h sf(a-bx+bh) * lim (a-bx) / h sf(a-bx) I dont know how to cancel h 4. Jun 3, 2008 rootX [tex] \frac{\sqrt{a-b(x+h)} - \sqrt{a-bx}}{h}$$

Try multiplying the numerator by
$$\frac{\sqrt{a-b(x+h)} + \sqrt{a-bx}}{\sqrt{a-b(x+h)} + \sqrt{a-bx}}$$

too hard to read what you wrote... >>

5. Jun 3, 2008

Asian Girl

too hard to read what you wrote... >>[/QUOTE]

Sorry, i dont know to use those symbols. Would you show me please?

6. Jun 3, 2008

konthelion

Basically, you multiply by the conjugate to remove the radicals in this situation.

For example, $$(\sqrt{c}+\sqrt{d})(\sqrt{c}-\sqrt{d})=c-d$$. where $$(\sqrt{c}-\sqrt{d})$$ is the conjugate of $$(\sqrt{c}+\sqrt{d})$$

7. Jun 3, 2008

Asian Girl

How do I use square roof symbol please?

8. Jun 3, 2008

konthelion

\sqrt{} you have to wrap it around "tex" and its end tag "/tex"

9. Jun 3, 2008

Asian Girl

That how I did but when I posted it did not show square roof's symbol.
Probably work's computer doesnt allow me to do it.

10. Jun 3, 2008

Kurdt

Staff Emeritus
Here is the thread that introduced latex:

If you click on any latex image you can see the code used. There are a couple of websites where you can practise your code linked to at the end of that thread. The forum has a drop down menu that looks like a capital sigma for adding things in latex.

11. Jun 3, 2008

Asian Girl

- b
------------------
sf (a-bx) + sf (a-bx)

so the domain of f ' is a-bx > 0

12. Jun 3, 2008

HallsofIvy

Staff Emeritus
"sf" is square root?

Yes, the derivative of $\sqrt{a-bx}$ is
$$\frac{-b}{2\sqrt{a-bx}}$$
just what you have.

13. Jun 3, 2008

konthelion

Solve for the domain of f, and f' in terms of x. Btw, it's a good thing to know that it is square root and not square "roof" =P

14. Jun 3, 2008

Asian Girl

my bad.... hehehehe ( I've been here just couple months)
Thanks so much all.

15. Jun 3, 2008

rootX

16. Jun 3, 2008

rootX

Last edited by a moderator: May 3, 2017