help with Derivative!
- Thread starter viet
- Start date
\(\displaystyle \L\\ f(x) = \frac{4x+5}{7x+4}\)
\(\displaystyle \L\\ f'(x) = \frac{(7x+4)(4x+5)' - (7x+4)'(4x+5)}{(7x+4)^2}\)
\(\displaystyle \L\\ f'(x) = \frac{(7x+4)(4) - (7)(4x+5)}{(7x+4)^2}\)
\(\displaystyle \L\\ f'(x) = \frac{(28x+16) - (28x+35)}{(7x+4)^2}\)
\(\displaystyle \L\\ f'(x) = \frac{(-19)}{(7x+4)^2}\)
\(\displaystyle \L\\ f'(3) = ?\)
\(\displaystyle \L\\ f'(x) = \frac{(7x+4)(4x+5)' - (7x+4)'(4x+5)}{(7x+4)^2}\)
\(\displaystyle \L\\ f'(x) = \frac{(7x+4)(4) - (7)(4x+5)}{(7x+4)^2}\)
\(\displaystyle \L\\ f'(x) = \frac{(28x+16) - (28x+35)}{(7x+4)^2}\)
\(\displaystyle \L\\ f'(x) = \frac{(-19)}{(7x+4)^2}\)
\(\displaystyle \L\\ f'(3) = ?\)
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.
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.
\(\displaystyle \L\\=-(7x+4)^{-2}(7)=\frac{-7}{(7x+4)^{2}}\)
\(\displaystyle \L\\=>(\frac{19}{7})(\frac{-7}{(7x+4)^{2}})\)=
\(\displaystyle \L\\\frac{-19}{(7x+4)^{2}}\)
First long divide 4x+5 by 7x+5. It's just one short step.
Code:
4/7
----------------
7x+4|4x+5
4x+16/7
--------------
19/7You 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).
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).
Do you know how to long divide?. You should by now if you're in calculus.
Anyway,
That gives us \(\displaystyle 1-\frac{6}{\sqrt{x}+3}\)
Now differentiate.
Just another approach.
Anyway,
Code:
1
----------------
sqrt(x)+3|sqrt(x)-3
sqrt(x)+3
---------------
-6That gives us \(\displaystyle 1-\frac{6}{\sqrt{x}+3}\)
Now differentiate.
Just another approach.