help with Derivative!

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viet

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Nov 19, 2005
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If f(x) = (4x+5)/(7x+4)
Find f'(3)

im not sure how to do this problem, so i use the Quotient rule. I got to be -1 which is wrong, can some help me with this?
 
\(\displaystyle \L\\ f(x) = \frac{4x+5}{7x+4}\)

\(\displaystyle \L\\ f'(x) = \frac{(7x+4)(4x+5)' - (7x+4)'(4x+5)}{(7x+4)^2}\)

\(\displaystyle \L\\ f'(x) = \frac{(7x+4)(4) - (7)(4x+5)}{(7x+4)^2}\)

\(\displaystyle \L\\ f'(x) = \frac{(28x+16) - (28x+35)}{(7x+4)^2}\)

\(\displaystyle \L\\ f'(x) = \frac{(-19)}{(7x+4)^2}\)

\(\displaystyle \L\\ f'(3) = ?\)
 
oh ok, i was a lil confused there, thanks for the help daon :D
 
There are different ways of tackling problems like these. If I may, here's another approach:

First long divide 4x+5 by 7x+5. It's just one short step.
Code: 
         4/7
        ----------------
    7x+4|4x+5
         4x+16/7
        --------------
               19/7
This gives: \(\displaystyle \L\\\frac{19}{7(7x+4)}+\frac{4}{7}\)

You know the derivative of 4/7 is 0, so throw it out.

You have:

\(\displaystyle \L\\\frac{19}{7}\frac{d}{dx}[\frac{1}{7x+4}]\)

Just take the derivative of \(\displaystyle \L\\(7x+4)^{-1}\) and multiply by 19/7.

\(\displaystyle \L\\=-(7x+4)^{-2}(7)=\frac{-7}{(7x+4)^{2}}\)

\(\displaystyle \L\\=>(\frac{19}{7})(\frac{-7}{(7x+4)^{2}})\)=

\(\displaystyle \L\\\frac{-19}{(7x+4)^{2}}\)
 
I stuck on this problem similar to the one above,

If f(x) = (sqrt(x)-3)/(sqrt(x)+3)

Find f'(x).

((sqrt(x)+3)(sqrt(x)-3)' - (sqrt(x)+3)'(sqrt(x)-3)) / (sqrt(x)+3)^2
=
((sqrt(x)+3)(sqrt(x))-(sqrt(x))(sqrt(x)-3)) / (sqrt(x)+3)^2

im stuck at this point, not sure if i should cancel out the sqrt(x).
 
You went wrong taking the deriviative of the square root

\(\displaystyle \L\\ \frac{d [sqrt{x}+3]}{dx} = \frac{1}{2sqrt{x}}\)
 
First, you could try long division to transform it to

\(\displaystyle \L\\1-\frac{6}{\sqrt{x}+3}\)

Now differentiate.
 
Do you know how to long divide?. You should by now if you're in calculus.

Anyway,

Code: 
              1
          ----------------
sqrt(x)+3|sqrt(x)-3
          sqrt(x)+3
              ---------------
                  -6

That gives us 1−6x+3\displaystyle 1-\frac{6}{\sqrt{x}+3}

Now differentiate.

Just another approach.
 
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