Homework Help: Find the derivative from first principles

Asian Girl · Jun 3, 2008

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)= [tex]\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

Defennder · Jun 3, 2008

Asian Girl said: ↑

lim h-->0 [tex]\frac{\sqrt{a-b(x+h)} - \sqrt{a-bx}}{h}[/tex]

lim h-->0 [tex]\frac{\sqrt{a-bx+bh-a+bx}}{h}[/tex]

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.

Asian Girl · Jun 3, 2008

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

rootX · Jun 3, 2008

[tex]
\frac{\sqrt{a-b(x+h)} - \sqrt{a-bx}}{h}
[/tex]

Try multiplying the numerator by
[tex]
\frac{\sqrt{a-b(x+h)} + \sqrt{a-bx}}{\sqrt{a-b(x+h)} + \sqrt{a-bx}}
[/tex]

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

Asian Girl · Jun 3, 2008

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

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

konthelion · Jun 3, 2008

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

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

Asian Girl · Jun 3, 2008

How do I use square roof symbol please?

konthelion · Jun 3, 2008

Asian Girl said: ↑

How do I use square roof symbol please?

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

Asian Girl · Jun 3, 2008

konthelion said: ↑

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

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.

Kurdt · Jun 3, 2008

Here is the thread that introduced latex:

https://www.physicsforums.com/showthread.php?t=8997

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.

Asian Girl · Jun 3, 2008

is this the right answer?

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

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

HallsofIvy · Jun 3, 2008

"sf" is square root?

Yes, the derivative of [itex]\sqrt{a-bx}[/itex] is
[tex]\frac{-b}{2\sqrt{a-bx}}[/tex]
just what you have.

konthelion · Jun 3, 2008

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

Asian Girl · Jun 3, 2008

konthelion said: ↑

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

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

rootX · Jun 3, 2008

Asian Girl said: ↑

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

something that would help ...
http://integrals.wolfram.com/index.jsp

rootX · Jun 3, 2008

oops.. that wasn't integration.. but there are some differentiation calcs too
like:
http://www.expertmathtutoring.com/Integral-Differential-Calculator.php
or
http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=mathcom&s1=calculus&s2=differentiate&s3=basic [Broken]