## anonymous 5 years ago find the derivative: (3x+1)^3/(1-3x)^4 I get the answer (3x+1)^2+(9x+21)/(1-3x)^5 is this correct?

1. anonymous

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

2. amistre64

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

3. anonymous

i have that much

4. amistre64

just working it out in me head :)

5. amistre64

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

6. anonymous

you lost me on that one

7. amistre64

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

8. anonymous

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

9. amistre64

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

10. amistre64

for simplicity: r = (1-3x) ; s = (3x+1) 3 r^3 s^2 (3r + 4s) ------------------ r^8

11. anonymous

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: $\frac{9 (1+3 x)^2}{(1-3 x)^4}+\frac{12 (1+3 x)^3}{(1-3 x)^5}$

12. amistre64

13. amistre64

might be right, but im lost on it :)

14. anonymous

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 $\frac{9 (1+3 x)^2}{(1-3 x)^4}+\frac{12 (1+3 x)^3}{(1-3 x)^5}$