Question

find the derivative:
(3x+1)^3/(1-3x)^4

I get the answer (3x+1)^2+(9x+21)/(1-3x)^5

is this correct?

Answer 1

the + between the 2 equations in the answer should not be there should be
(3x+1)^2 (9x+21)/(1-3x)^5

Answer 2

[ 9 (1-3x)^4 (3x+1)^2 ] - [ -12 (3x+1)^3 (1-3x)^3 ]
----------------------------------------------
(1-3x)^8

Answer 3

i have that much

Answer 4

just working it out in me head :)

Answer 5

9bt' - (-12b't) 9bt' + 12b't
------------ = ------------
b^2 b^2

the (+) is good

Answer 6

you lost me on that one

Answer 7

we get 2 terms up top that are made up of multiplication.

the first term has a (+)9 as a constant, and the 2nd term has a (-)12 as a constant.

9 --12 = 9+12

Answer 8

ok I got that. I guess the way it was written confussed me

Answer 9

probably, I was just trying to clean it up for my eyes :)

Answer 10

for simplicity:

r = (1-3x) ; s = (3x+1)

3 r^3 s^2 (3r + 4s)
------------------
r^8

Answer 11

Let f(x) and g(x) be the Numerator and the Denominator of the given fraction respectively.

The derivative of the fraction in terms of the above functions is:\[\frac{f'[x]}{g[x]}-\frac{f[x] g'[x]}{g[x]^2} \]
Plug in the function values and their associated derivatives and you should get:
\[\frac{9 (1+3 x)^2}{(1-3 x)^4}+\frac{12 (1+3 x)^3}{(1-3 x)^5} \]

Answer 12

rob: I dont follow that.....

Answer 13

might be right, but im lost on it :)

Answer 14

To start from the beginning. mom wants to know if the derivative of
\[{(3x+1)^3 \over {(1-3x)^4}} \]
is equal to:
\[(3 x+1)^2+\frac{(9 x+21)}{(1-3 x)^5} \]
The answer is no. The derivative is:
\[\frac{9 (1+3 x)^2}{(1-3 x)^4}+\frac{12 (1+3 x)^3}{(1-3 x)^5} \]
Let
\[\frac{(3 x+1)^3}{(1-3 x)^4}=\frac{f(x)}{g(x)} \]
The derivative of f(x)/g(x) is
\[\frac{f'(x)}{g(x)}-\frac{f(x) g'(x)}{g(x)^2} \]
\[f(x) = (1+3x)^3, f'(x) = 9(1+3x)^2 \]
\[g(x) = (1-3x)^4, g'(x) = -12(1-3x)^3\]
Plug in the values for f(x), f'(x), g(x) and g'(x) into the derivative of the f(x)/g(x) and one should end up with the equivalent of the derivative.
\[\frac{9 (1+3 x)^2}{(1-3 x)^4}-\frac{(1+3 x)^3 \left(-12 (1-3 x)^3\right)}{\left((1-3 x)^4\right)^2} \]
or
\[\frac{9 (1+3 x)^2}{(1-3 x)^4}+\frac{12 (1+3 x)^3}{(1-3 x)^5} \]