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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=(2x5)^3(1x^4)^2
working it out I get:
(2x5)^3 [2(1x^4)(4x^3)] + (1x^4)^2 [3(2x5)^2 (2)]
which is
(2x5)^3[8x^3(1x^4)] +(1x^4)^2[6(2x5)^2]
The online guide says to now factor and ends up with
2(2x5)^2(1x^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
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=(2x5)^3(1x^4)^2 working it out I get: (2x5)^3 [2(1x^4)(4x^3)] + (1x^4)^2 [3(2x5)^2 (2)] which is (2x5)^3[8x^3(1x^4)] +(1x^4)^2[6(2x5)^2] The online guide says to now factor and ends up with 2(2x5)^2(1x^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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[y=(2x5)^3(1x^4)^2\]And you're trying to take the derivative of this using the product rule  \(f'g + g'f\)?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1you did: \[y=(2x5)^3(1x^4)^2\] \[(2x5)^3 [2(1x^4)(4x^3)] + (1x^4)^2 [3(2x5)^2 (2)]\] \[(2x5)^3[8x^3(1x^4)] +(1x^4)^2[6(2x5)^2]\] and they want \[2(2x5)^2(1x^4)[11x^4+20x^3+3]\] right?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1make it easy for yourself write (2x5) as A and (1x^4) as B then try

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1you have \[A^3[8x^3B] +B^2[6A^2] \] they want \[2A^2B[11x^4+20x^3+3]\] something's gotta give!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't understand. I cannot see the relation

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0\(\large\color{black}{ \displaystyle y=(2x5)^3(1x^4)^2 }\) \(\large\color{black}{ \displaystyle\ln y= \ln\left[(2x5)^3(1x^4)^2\right] }\) \(\large\color{black}{ \displaystyle \ln y=\ln\left[(2x5)^3\right]+\ln\left[(1x^4)^2\right] }\) \(\large\color{black}{ \displaystyle \ln y=3\ln\left[2x5\right]+2\ln\left[1x^4\right] }\) \(\large\color{black}{ \displaystyle \frac{y'}{y}=3\cdot \frac{2}{2x5}+2\cdot\frac{4x^3}{1x^4} }\) \(\large\color{black}{ \displaystyle y'=y\left(\frac{6}{2x5}+\frac{8x^3}{1x^4}\right) }\) \(\large\color{black}{ \displaystyle y'=(2x5)^3(1x^4)^2\left(\frac{6}{2x5}+\frac{8x^3}{1x^4}\right) }\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[y=(2x5)^3(1x^4)^2\]\[y' = 3\color{blue}{(2x5)^2}(2)\cdot \color{red}{ (1x^4)^2} +2\color{red}{(1x^4)}(4x^3)\cdot \color{blue}{(2x5)^3}\]\[y'= \color{blue}{(2x5)^2}\color{red}{(1x^4)}\left[6(1x^4)+2(4x^3)(2x5)^2\right]\]\[y'=\color{blue}{(2x5)^2}\color{red}{(1x^4)}\left[66x^4+8x^3(4x^220x+25)\right]\]\[y'=\color{blue}{(2x5)^2}\color{red}{(1x^4)}\left[66x^4+32x^5160^4+200x^3\right]\]\[\boxed{y'=\color{blue}{(2x5)^2}\color{red}{(1x^4)}\left[6166x^4+32x^5+200x^3\right]}\] @SolomonZelman check my work lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think I forgot a negative somewhere in there.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.04x^3 is a negative chain

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep. I spotted it too

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0oh, I left it out too.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0\(\large\color{black}{ \displaystyle y=(2x5)^3(1x^4)^2 }\) \(\large\color{black}{ \displaystyle\ln y= \ln\left[(2x5)^3(1x^4)^2\right] }\) \(\large\color{black}{ \displaystyle \ln y=\ln\left[(2x5)^3\right]+\ln\left[(1x^4)^2\right] }\) \(\large\color{black}{ \displaystyle \ln y=3\ln\left[2x5\right]+2\ln\left[1x^4\right] }\) < ☼ CORRECTION ☼ > \(\large\color{black}{ \displaystyle \frac{y'}{y}=3\cdot \frac{2}{2x5}+2\cdot\frac{4x^3}{1x^4} }\) \(\large\color{black}{ \displaystyle y'=y\left(\frac{6}{2x5}\frac{8x^3}{1x^4}\right) }\) \(\large\color{black}{ \displaystyle y'=(2x5)^3(1x^4)^2\left(\frac{6}{2x5}\frac{8x^3}{1x^4}\right) }\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[y' = 3\color{blue}{(2x5)^2}(2)\cdot \color{red}{ (1x^4)^2} +2\color{red}{(1x^4)}(4x^3)\cdot \color{blue}{(2x5)^3}\]\[y'= \color{blue}{(2x5)^2}\color{red}{(1x^4)}\left[6(1x^4)+2(4x^3)(2x5)^2\right]\]\[y'=\color{blue}{(2x5)^2}\color{red}{(1x^4)}\left[66x^48x^3(4x^220x+25)\right]\]\[y'=\color{blue}{(2x5)^2}\color{red}{(1x^4)}\left[66x^432x^5+160^4200x^3\right]\]\[\boxed{y'=\color{blue}{(2x5)^2}\color{red}{(1x^4)}\left[6154x^432x^5200x^3\right]}\] Theres my correction.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\boxed{y'=\color{blue}{(2x5)^2}\color{red}{(1x^4)}\left[6+154x^432x^5200x^3\right]}\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1@mthompson440 i meant this by the suggestion: write (2x5) as A and (1x^4) as B with the simplifications, they want \[2A^2B[11x^4+20x^3+3]\] you have \(A^3(8x^3B) +B^2[6A^2]\) \(= 2A^3B(4x^3) + 3A^2 B^2\) \(= 2A^2B(A(8x^3) + 6B)\) \(= 2A^2B(4(2x5)x^3+3(1x^4))\) \(= 2A^2B(8x^4+20x^3+33x^4))\) \(= 2A^2B(11x^4+20x^3+3))\) 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
 one year ago
Best ResponseYou've already chosen the best response.1Where is the koala bear?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1@Jhannybean \[y'= \color{blue}{(2x5)^2}\color{red}{(1x^4)}\left[6(1x^4)+2(4x^3)(2x5)^\color{green}{1}\right]\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I put the correction in green because you already used the prettier colors

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Refer to the attachment from Mathematica v9.
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