anonymous
  • anonymous
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
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Jhannybean
  • Jhannybean
\[y=(2x-5)^3(1-x^4)^2\]And you're trying to take the derivative of this using the product rule - \(f'g + g'f\)?
anonymous
  • anonymous
yes
IrishBoy123
  • 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?

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anonymous
  • anonymous
yes
IrishBoy123
  • IrishBoy123
make it easy for yourself write (2x-5) as A and (1-x^4) as B then try
IrishBoy123
  • IrishBoy123
you have \[A^3[-8x^3B] +B^2[6A^2] \] they want \[2A^2B[-11x^4+20x^3+3]\] something's gotta give!!
anonymous
  • anonymous
I don't understand. I cannot see the relation
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle y=(2x-5)^3(1-x^4)^2 }\) \(\large\color{black}{ \displaystyle\ln y= \ln\left[(2x-5)^3(1-x^4)^2\right] }\) \(\large\color{black}{ \displaystyle \ln y=\ln\left[(2x-5)^3\right]+\ln\left[(1-x^4)^2\right] }\) \(\large\color{black}{ \displaystyle \ln y=3\ln\left[2x-5\right]+2\ln\left[1-x^4\right] }\) \(\large\color{black}{ \displaystyle \frac{y'}{y}=3\cdot \frac{2}{2x-5}+2\cdot\frac{4x^3}{1-x^4} }\) \(\large\color{black}{ \displaystyle y'=y\left(\frac{6}{2x-5}+\frac{8x^3}{1-x^4}\right) }\) \(\large\color{black}{ \displaystyle y'=(2x-5)^3(1-x^4)^2\left(\frac{6}{2x-5}+\frac{8x^3}{1-x^4}\right) }\)
Jhannybean
  • 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
Jhannybean
  • Jhannybean
I think I forgot a negative somewhere in there.
SolomonZelman
  • SolomonZelman
-4x^3 is a negative chain
Jhannybean
  • Jhannybean
Yep. I spotted it too
SolomonZelman
  • SolomonZelman
oh, I left it out too.
SolomonZelman
  • 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 ☼ > \(\large\color{black}{ \displaystyle \frac{y'}{y}=3\cdot \frac{2}{2x-5}+2\cdot\frac{-4x^3}{1-x^4} }\) \(\large\color{black}{ \displaystyle y'=y\left(\frac{6}{2x-5}-\frac{8x^3}{1-x^4}\right) }\) \(\large\color{black}{ \displaystyle y'=(2x-5)^3(1-x^4)^2\left(\frac{6}{2x-5}-\frac{8x^3}{1-x^4}\right) }\)
Jhannybean
  • 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.
Jhannybean
  • Jhannybean
\[\boxed{y'=\color{blue}{(2x-5)^2}\color{red}{(1-x^4)}\left[6+154x^4-32x^5-200x^3\right]}\]
IrishBoy123
  • 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
freckles
  • freckles
Where is the koala bear?
freckles
  • freckles
@Jhannybean \[y'= \color{blue}{(2x-5)^2}\color{red}{(1-x^4)}\left[6(1-x^4)+2(-4x^3)(2x-5)^\color{green}{1}\right]\]
freckles
  • freckles
I put the correction in green because you already used the prettier colors
anonymous
  • anonymous
Refer to the attachment from Mathematica v9.
1 Attachment

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