Question

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

Answer 1

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

Answer 2

yes

Answer 3

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?

Answer 4

yes

Answer 5

make it easy for yourself

write (2x-5) as A and (1-x^4) as B

then try

Answer 6

you have
\[A^3[-8x^3B] +B^2[6A^2] \]

they want
\[2A^2B[-11x^4+20x^3+3]\]

something's gotta give!!

Answer 7

I don't understand. I cannot see the relation

Answer 8

\(\large\color{black}{ \displaystyle y=(2x-5)^3(1-x^4)^2 }\)

\(\large\color{black}{ \displaystyle\ln y= \ln\left[(2x-5)^3(1-x^4)^2\right] }\)

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

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

\(\large\color{black}{ \displaystyle \frac{y'}{y}=3\cdot \frac{2}{2x-5}+2\cdot\frac{4x^3}{1-x^4} }\)

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

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

Answer 9

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

Answer 10

I think I forgot a negative somewhere in there.

Answer 11

-4x^3 is a negative chain

Answer 12

Yep. I spotted it too

Answer 13

oh, I left it out too.

Answer 14

\(\large\color{black}{ \displaystyle y=(2x-5)^3(1-x^4)^2 }\)

\(\large\color{black}{ \displaystyle\ln y= \ln\left[(2x-5)^3(1-x^4)^2\right] }\)

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

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


< ☼ CORRECTION ☼ >


\(\large\color{black}{ \displaystyle \frac{y'}{y}=3\cdot \frac{2}{2x-5}+2\cdot\frac{-4x^3}{1-x^4} }\)

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

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

Answer 15

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

Answer 16

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

Answer 17

@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

Answer 18

Where is the koala bear?

Answer 19

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

Answer 20

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

Answer 21

Refer to the attachment from Mathematica v9.