Homework Help: Find the derivative from first principles

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.

 

[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... >>

 

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]

 

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

 

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.

 

"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.

 

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

 

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

 

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]