Double prime?

[viet]

the problem said to fine f ' (2), f '' (2), and f ''' (2)
f(x) = (1-3x)/(1+3x)

i used the quotient rule to fine f '(x):
f '(x) = (1+3x)(1-3x)' - (1+3x)'(1-3x) / (1+3x)^2
f '(x) = (1+3x)(-3) - (3)(1+3x) / (1+3x)^2
f '(x) = (-3-9x) - (3-9x) / (1+3x)^2
f '(x) = -6 / (1+3x)^2

f '(2) = -6 / (1+3(2))^2

the next part is to find f ''(2) and f '''(2)
im not sure how to do find that can someone help me with this? thanks in advance

 

[royhaas]

Apply the same rules to f' to find f'', and so on.

 

[viet]

so use the quotient rule again and apply it to f '(x) = -6 / (1+3x)^2 ?
sofar i got to:
f ''(x) = ((1+3x)^2)(-6)' - ((1+3x)^2)'(-6) / ((1+3x)^2)^2

 

[soroban]

Hello, viet!

$\displaystyle f(x)\;=\;\frac{1\,+\,3x}{1\,-\,3x}\;$ Find: $\displaystyle f'(2),\;f''(2),\;f'''(-2)$

I used the quotient rule to find $\displaystyle f'(x)$:
$\displaystyle f'(x)\:=\:\frac{(1\,+\,3x)(-3)\,-\,(3)(1+3x)}{(1\,+\,3x)^2}$
$\displaystyle f'(x)\:=\:\frac{(-3\,-\,9x)\,-\,(3\,-\,9x)}{(1\,+\,3x)^2}$
$\displaystyle f'(x)\:=\:\frac{-6}{(1\,+\,3x)^2}$

$\displaystyle f'(2)\:=\:\frac{-6}{(1\,+\,3\cdot2)^2}\:=\:-\frac{6}{49}$

You have: $\displaystyle \,f'(x)\:=\:-6(1\,+\,3x)^{-2}$

Then: $\displaystyle \,f''(x)\:=\:12(1\,+\,3x)^{-3}(3)\;=\;36(1\,+\,3x)^{-3}\:=\:\frac{36}{(1\,+\,3x)^3}$

Hence: $\displaystyle \,f''(2)\:=\:\frac{36}{(1\,+\,3\cdot2)^3} \:= \;\frac{36}{343}$


Since we have: $\displaystyle \,f''(x)\:=\;36(1\,+\,3x)^{-3}$,
$\displaystyle \;\;$ then: $\displaystyle \,f'''(x)\;=\;-108(1\,+\,3x)^{-4}\cdot3\;=\;\frac{-324}{(1\,+\,3x)^4}$

Hence: \(\displaystyle \,f'''(2)\:=\:\frac{-324}{(1\,+\,3\cdot2)^4}\;=\;-\frac{324}{2401\)

 

[viet]

oh ok. thanks the explanation!