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

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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
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\[y=(2x-5)^3(1-x^4)^2\]And you're trying to take the derivative of this using the product rule - \(f'g + g'f\)?
yes
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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yes
make it easy for yourself write (2x-5) as A and (1-x^4) as B then try
you have \[A^3[-8x^3B] +B^2[6A^2] \] they want \[2A^2B[-11x^4+20x^3+3]\] something's gotta give!!
I don't understand. I cannot see the relation
\(\large\color{black}{ \displaystyle y=(2x-5)^3(1-x^4)^2 }\) \(\large\color{black}{ \displaystyle\ln y= \ln\left[(2x-5)^3(1-x^4)^2\right] }\) \(\large\color{black}{ \displaystyle \ln y=\ln\left[(2x-5)^3\right]+\ln\left[(1-x^4)^2\right] }\) \(\large\color{black}{ \displaystyle \ln y=3\ln\left[2x-5\right]+2\ln\left[1-x^4\right] }\) \(\large\color{black}{ \displaystyle \frac{y'}{y}=3\cdot \frac{2}{2x-5}+2\cdot\frac{4x^3}{1-x^4} }\) \(\large\color{black}{ \displaystyle y'=y\left(\frac{6}{2x-5}+\frac{8x^3}{1-x^4}\right) }\) \(\large\color{black}{ \displaystyle y'=(2x-5)^3(1-x^4)^2\left(\frac{6}{2x-5}+\frac{8x^3}{1-x^4}\right) }\)
\[y=(2x-5)^3(1-x^4)^2\]\[y' = 3\color{blue}{(2x-5)^2}(2)\cdot \color{red}{ (1-x^4)^2} +2\color{red}{(1-x^4)}(4x^3)\cdot \color{blue}{(2x-5)^3}\]\[y'= \color{blue}{(2x-5)^2}\color{red}{(1-x^4)}\left[6(1-x^4)+2(4x^3)(2x-5)^2\right]\]\[y'=\color{blue}{(2x-5)^2}\color{red}{(1-x^4)}\left[6-6x^4+8x^3(4x^2-20x+25)\right]\]\[y'=\color{blue}{(2x-5)^2}\color{red}{(1-x^4)}\left[6-6x^4+32x^5-160^4+200x^3\right]\]\[\boxed{y'=\color{blue}{(2x-5)^2}\color{red}{(1-x^4)}\left[6-166x^4+32x^5+200x^3\right]}\] @SolomonZelman check my work lol
I think I forgot a negative somewhere in there.
-4x^3 is a negative chain
Yep. I spotted it too
oh, I left it out too.
\(\large\color{black}{ \displaystyle y=(2x-5)^3(1-x^4)^2 }\) \(\large\color{black}{ \displaystyle\ln y= \ln\left[(2x-5)^3(1-x^4)^2\right] }\) \(\large\color{black}{ \displaystyle \ln y=\ln\left[(2x-5)^3\right]+\ln\left[(1-x^4)^2\right] }\) \(\large\color{black}{ \displaystyle \ln y=3\ln\left[2x-5\right]+2\ln\left[1-x^4\right] }\) < ☼ CORRECTION ☼ > \(\large\color{black}{ \displaystyle \frac{y'}{y}=3\cdot \frac{2}{2x-5}+2\cdot\frac{-4x^3}{1-x^4} }\) \(\large\color{black}{ \displaystyle y'=y\left(\frac{6}{2x-5}-\frac{8x^3}{1-x^4}\right) }\) \(\large\color{black}{ \displaystyle y'=(2x-5)^3(1-x^4)^2\left(\frac{6}{2x-5}-\frac{8x^3}{1-x^4}\right) }\)
\[y' = 3\color{blue}{(2x-5)^2}(2)\cdot \color{red}{ (1-x^4)^2} +2\color{red}{(1-x^4)}(4x^3)\cdot \color{blue}{(2x-5)^3}\]\[y'= \color{blue}{(2x-5)^2}\color{red}{(1-x^4)}\left[6(1-x^4)+2(-4x^3)(2x-5)^2\right]\]\[y'=\color{blue}{(2x-5)^2}\color{red}{(1-x^4)}\left[6-6x^4-8x^3(4x^2-20x+25)\right]\]\[y'=\color{blue}{(2x-5)^2}\color{red}{(1-x^4)}\left[6-6x^4-32x^5+160^4-200x^3\right]\]\[\boxed{y'=\color{blue}{(2x-5)^2}\color{red}{(1-x^4)}\left[6-154x^4-32x^5-200x^3\right]}\] Theres my correction.
\[\boxed{y'=\color{blue}{(2x-5)^2}\color{red}{(1-x^4)}\left[6+154x^4-32x^5-200x^3\right]}\]
@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
Where is the koala bear?
@Jhannybean \[y'= \color{blue}{(2x-5)^2}\color{red}{(1-x^4)}\left[6(1-x^4)+2(-4x^3)(2x-5)^\color{green}{1}\right]\]
I put the correction in green because you already used the prettier colors
Refer to the attachment from Mathematica v9.
1 Attachment

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