Find the derivative from first principles

[Asian Girl]

Homework Statement

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 ' ?

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]

lim h-->0 $$\frac{\sqrt{a-b(x+h)} - \sqrt{a-bx}}{h}$$

lim h-->0 $$\frac{\sqrt{a-bx+bh-a+bx}}{h}$$

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]

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]

$$\frac{\sqrt{a-b(x+h)} - \sqrt{a-bx}}{h}$$

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... >>

 

[Asian Girl]

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


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

 

[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 $$(\sqrt{c}+\sqrt{d})$$

 

[Asian Girl]

How do I use square roof symbol please?

 

[konthelion]

How do I use square roof symbol please?

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

 

[Asian Girl]

\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]

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]

is this the right answer?

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

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

 

[HallsofIvy]

"sf" is square root?

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

 

[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

 

[Asian Girl]

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]

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]

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]