## anonymous one year ago I have a calculus / derivative problem that I am unable to understand how the last step is worked out. The problem is y=(2x-5)^3(1-x^4)^2 working it out I get: (2x-5)^3 [2(1-x^4)(-4x^3)] + (1-x^4)^2 [3(2x-5)^2 (2)] which is (2x-5)^3[-8x^3(1-x^4)] +(1-x^4)^2[6(2x-5)^2] The online guide says to now factor and ends up with 2(2x-5)^2(1-x^4)[-11x^4+20x^3+3] I am unable to see what was factored and how the final answer was arrived at. Any help is appreciated. Straight answers are best. Asking me to try and guess is frustrating to me. Thanks

1. Jhannybean

$y=(2x-5)^3(1-x^4)^2$And you're trying to take the derivative of this using the product rule - $$f'g + g'f$$?

2. anonymous

yes

3. IrishBoy123

you did: $y=(2x-5)^3(1-x^4)^2$ $(2x-5)^3 [2(1-x^4)(-4x^3)] + (1-x^4)^2 [3(2x-5)^2 (2)]$ $(2x-5)^3[-8x^3(1-x^4)] +(1-x^4)^2[6(2x-5)^2]$ and they want $2(2x-5)^2(1-x^4)[-11x^4+20x^3+3]$ right?

4. anonymous

yes

5. IrishBoy123

make it easy for yourself write (2x-5) as A and (1-x^4) as B then try

6. IrishBoy123

you have $A^3[-8x^3B] +B^2[6A^2]$ they want $2A^2B[-11x^4+20x^3+3]$ something's gotta give!!

7. anonymous

I don't understand. I cannot see the relation

8. SolomonZelman

$$\large\color{black}{ \displaystyle y=(2x-5)^3(1-x^4)^2 }$$ $$\large\color{black}{ \displaystyle\ln y= \ln\left[(2x-5)^3(1-x^4)^2\right] }$$ $$\large\color{black}{ \displaystyle \ln y=\ln\left[(2x-5)^3\right]+\ln\left[(1-x^4)^2\right] }$$ $$\large\color{black}{ \displaystyle \ln y=3\ln\left[2x-5\right]+2\ln\left[1-x^4\right] }$$ $$\large\color{black}{ \displaystyle \frac{y'}{y}=3\cdot \frac{2}{2x-5}+2\cdot\frac{4x^3}{1-x^4} }$$ $$\large\color{black}{ \displaystyle y'=y\left(\frac{6}{2x-5}+\frac{8x^3}{1-x^4}\right) }$$ $$\large\color{black}{ \displaystyle y'=(2x-5)^3(1-x^4)^2\left(\frac{6}{2x-5}+\frac{8x^3}{1-x^4}\right) }$$

9. Jhannybean

$y=(2x-5)^3(1-x^4)^2$$y' = 3\color{blue}{(2x-5)^2}(2)\cdot \color{red}{ (1-x^4)^2} +2\color{red}{(1-x^4)}(4x^3)\cdot \color{blue}{(2x-5)^3}$$y'= \color{blue}{(2x-5)^2}\color{red}{(1-x^4)}\left[6(1-x^4)+2(4x^3)(2x-5)^2\right]$$y'=\color{blue}{(2x-5)^2}\color{red}{(1-x^4)}\left[6-6x^4+8x^3(4x^2-20x+25)\right]$$y'=\color{blue}{(2x-5)^2}\color{red}{(1-x^4)}\left[6-6x^4+32x^5-160^4+200x^3\right]$$\boxed{y'=\color{blue}{(2x-5)^2}\color{red}{(1-x^4)}\left[6-166x^4+32x^5+200x^3\right]}$ @SolomonZelman check my work lol

10. Jhannybean

I think I forgot a negative somewhere in there.

11. SolomonZelman

-4x^3 is a negative chain

12. Jhannybean

Yep. I spotted it too

13. SolomonZelman

oh, I left it out too.

14. SolomonZelman

$$\large\color{black}{ \displaystyle y=(2x-5)^3(1-x^4)^2 }$$ $$\large\color{black}{ \displaystyle\ln y= \ln\left[(2x-5)^3(1-x^4)^2\right] }$$ $$\large\color{black}{ \displaystyle \ln y=\ln\left[(2x-5)^3\right]+\ln\left[(1-x^4)^2\right] }$$ $$\large\color{black}{ \displaystyle \ln y=3\ln\left[2x-5\right]+2\ln\left[1-x^4\right] }$$ < ☼ CORRECTION ☼ > $$\large\color{black}{ \displaystyle \frac{y'}{y}=3\cdot \frac{2}{2x-5}+2\cdot\frac{-4x^3}{1-x^4} }$$ $$\large\color{black}{ \displaystyle y'=y\left(\frac{6}{2x-5}-\frac{8x^3}{1-x^4}\right) }$$ $$\large\color{black}{ \displaystyle y'=(2x-5)^3(1-x^4)^2\left(\frac{6}{2x-5}-\frac{8x^3}{1-x^4}\right) }$$

15. Jhannybean

$y' = 3\color{blue}{(2x-5)^2}(2)\cdot \color{red}{ (1-x^4)^2} +2\color{red}{(1-x^4)}(4x^3)\cdot \color{blue}{(2x-5)^3}$$y'= \color{blue}{(2x-5)^2}\color{red}{(1-x^4)}\left[6(1-x^4)+2(-4x^3)(2x-5)^2\right]$$y'=\color{blue}{(2x-5)^2}\color{red}{(1-x^4)}\left[6-6x^4-8x^3(4x^2-20x+25)\right]$$y'=\color{blue}{(2x-5)^2}\color{red}{(1-x^4)}\left[6-6x^4-32x^5+160^4-200x^3\right]$$\boxed{y'=\color{blue}{(2x-5)^2}\color{red}{(1-x^4)}\left[6-154x^4-32x^5-200x^3\right]}$ Theres my correction.

16. Jhannybean

$\boxed{y'=\color{blue}{(2x-5)^2}\color{red}{(1-x^4)}\left[6+154x^4-32x^5-200x^3\right]}$

17. IrishBoy123

@mthompson440 i meant this by the suggestion: write (2x-5) as A and (1-x^4) as B with the simplifications, they want $2A^2B[-11x^4+20x^3+3]$ you have $$A^3(-8x^3B) +B^2[6A^2]$$ $$= 2A^3B(-4x^3) + 3A^2 B^2$$ $$= 2A^2B(A(-8x^3) + 6B)$$ $$= 2A^2B(-4(2x-5)x^3+3(1-x^4))$$ $$= 2A^2B(-8x^4+20x^3+3-3x^4))$$ $$= 2A^2B(-11x^4+20x^3+3))$$ there's no silver bullet for this kind of mess. just look up at this thread! that was just a suggestion as to how to make life easier. i am sure you can think of your own :p

18. freckles

Where is the koala bear?

19. freckles

@Jhannybean $y'= \color{blue}{(2x-5)^2}\color{red}{(1-x^4)}\left[6(1-x^4)+2(-4x^3)(2x-5)^\color{green}{1}\right]$

20. freckles

I put the correction in green because you already used the prettier colors

21. anonymous

Refer to the attachment from Mathematica v9.