Question

h(t)=(t+1)^2/3(2t-1)^3
h'(t)=(t+1)^2/3*3(2t^2-1)^2*4t+(2t^2-1)^3*2/3(t+1)^-1/3
I am stuck right here please help! Using the chain rule I have done ok. It's the Algebra that is killing me!

Answer 1

There's not much else you can do here. I guess you could try to simplify, but I don't think any amount of algebra will actually simplify this. It will just rewrite it in another form really.

Answer 2

You can factor

Answer 3

yeah I guess you could factor out a t+1 term

Answer 4

if you factor out say (t+1)^(2/3), then you divide (t+1)^2/3*3(2t^2-1)^2*4t by (t+1)^(2/3)

and you also divide (2t^2-1)^3*2/3(t+1)^-1/3 by (t+1)^(2/3)

Answer 5

But see you would have to factor 2/3(t+1)^1/3 and (2t^2-1)^2

Answer 6

I'm hoping Perl is got a nice solid answer...

Answer 7

one moment, i might try something different

Answer 8

I see what Jim was sort of saying. What program do you use?

Answer 9

i noticed in your question, a square appears in the derivative.
is
h(t)=(t+1)^2/3(2t-1)^3
or
h(t)=(t+1)^2/3(2t^ `2`-1)^3

Answer 10

using the chain rule it is converted to (4t)

Answer 11

oh that's a good point @perl, I didn't notice that til just now

Answer 12

So what does the square do?

Answer 13

I'm sorry I am missing that point

Answer 14

h(t)=(t+1?13(2t2 - 1)3 =>
h'(t)= (t+1?/3 . 3(2t2 - 1)2 . 4t+(2t2 -1)3 . +1)-113 = +1)-113(2t2 - 1)2[18t(t+1)+(2t2- 1)]
=i(t+1)-113(2t2- 1)2(20t2+18t-1)

Answer 15

I'm sorry that came out ugly

Answer 16

$$
\large {h(t) = (t+1)^\frac{2}{3}(2t\color{red}{^2}-1)^3
\\
\large { h'(t)=(t+1)^{2/3}\cdot 3(2t^2-1)^2~4t+\frac{2}{3}(t+1)^{-1/3} (2t^2-1)^3\cdot \\

=\frac{2}{3}(2t^2-1)^2(t+1)^{-\frac{1}{3}}[~(t+1)^{\frac{2}{3}+\frac{1}{3} }\cdot 2\cdot 9 t +(2t^2 -1)]
}}

$$

Answer 17

any questions so far?

Answer 18

I'm following you!

Answer 19

Well why are you adding 2/3+1/3?

Answer 20

try distributing

Answer 21

if we go backwards and distribute we have:
\(\color{blue}{\text{Originally Posted by}}\)
\[
{

\frac{2}{3}(2t^2-1)^2(t+1)^{-\frac{1}{3}}[~(t+1)^{\frac{2}{3}+\frac{1}{3} }\cdot 2\cdot 9 t +(2t^2 -1)]\\\text{ }\\
=\\\text { }\\
\frac{2}{3}(2t^2-1)^2(t+1)^{-\frac{1}{3}}\cdot ~(t+1)^{\frac{2}{3}+\frac{1}{3} }\cdot 2\cdot 9 t +\frac{2}{3}(2t^2-1)^2(t+1)^{-\frac{1}{3}}(2t^2 -1)]
\\\text { }\\=\\\text { }\\
\frac{2}{3}\cdot 2\cdot 9 t~(2t^2-1)^2 ~(t+1)^{-\frac{1}{3} +\frac{2}{3}+\frac{1}{3} } +\frac{2}{3}(2t^2-1)^{2+1}(t+1)^{-\frac{1}{3}}

\\\text { }\\=\\\text { }\\
3\cdot 4t~(2t^2-1)^2 ~(t+1)^{\frac{2}{3} } +\frac{2}{3}(2t^2-1)^{3}(t+1)^{-\frac{1}{3}}
\\\text { }\\\text{so they are equivalent }\\\text { }\\


}




\]
\(\color{blue}{\text{End of Quote}}\)

Answer 22

you can only use the chain rule on the denominator