Can you help???

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mathhelp1a

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Differentiate and simplify

(2t-1) ^ 3 (t^2+1)^5

is it (2t-1)^3 [5(t^2+1)^4] + (t^2+1)^5 [3(2t-1)]

then what you i do
 
mathhelp1a said:
Differentiate and simplify

(2t-1) ^ 3 (t^2+1)^5

is it (2t-1)^3 [5(t^2+1)^4] + (t^2+1)^5 [3(2t-1)]
Click to expand...

Almost, but not quite.

You're applied the product rule correctly (to get [sup:3mvei4t2]d[/sup:3mvei4t2]/[sub:3mvei4t2]dt[/sub:3mvei4t2] (2t-1) ^ 3 (t^2+1)^5 = (2t-1) ^ 3 ([sup:3mvei4t2]d[/sup:3mvei4t2]/[sub:3mvei4t2]dt[/sub:3mvei4t2] (t^2+1)^5) + (2t-1) ^ 3
([sup:3mvei4t2]d[/sup:3mvei4t2]/[sub:3mvei4t2]dt[/sub:3mvei4t2] (t^2+1)^5)

However, you haven't applied the chain rule correctly to find [sup:3mvei4t2]d[/sup:3mvei4t2]/[sub:3mvei4t2]dt[/sub:3mvei4t2] (2t-1) ^ 3 and [sup:3mvei4t2]d[/sup:3mvei4t2]/[sub:3mvei4t2]dt[/sub:3mvei4t2] (t^2+1)^5

According to the chain rule, [sup:3mvei4t2]d[/sup:3mvei4t2]/[sub:3mvei4t2]dt[/sub:3mvei4t2] f(g(t)) = f'(g(t)) [sup:3mvei4t2]d[/sup:3mvei4t2]/[sub:3mvei4t2]dt[/sub:3mvei4t2] g(t). So, for example, [sup:3mvei4t2]d[/sup:3mvei4t2]/[sub:3mvei4t2]dt[/sub:3mvei4t2] (t^3+7t)^9 = 9(t^3+7t)^8 [sup:3mvei4t2]d[/sup:3mvei4t2]/[sub:3mvei4t2]dt[/sub:3mvei4t2] (t^3+7t), not just 9(t^3+7t)^8.

then what you i do
Click to expand...

"Simplify".

For example, your answer will be a sum of two terms, these have a common factor of (2t-1)^2 (t^2+1)^4. You could take this out. Then, what's left of the sum can be multiplied out and simplified.
 
You were closer the very first time.

I'll do a similar example...

\(\displaystyle \frac{d}{dx}(3x+4)^6(x+x^{-1})^9\)
\(\displaystyle =(3x+4)^6\left[\frac{d}{dx}(x+x^{-1})^9\right] + (x+x^{-1})^9\left[\frac{d}{dx}(3x+4)^6\right]\)
\(\displaystyle =(3x+4)^6\left[9(x+x^{-1})^8\left(\frac{d}{dx}(x+x^{-1})\right)\right] + (x+x^{-1})^9\left[6(3x+4)^5\left(\frac{d}{dx}(3x+4)\right)\right]\)
\(\displaystyle =(3x+4)^6\left[9(x+x^{-1})^8\left(1-x^{-2}\right)\right] + (x+x^{-1})^9\left[6(3x+4)^5\left(3\right)\right]\)

Now it's all differentiated, I'll start to simplify...

\(\displaystyle (3x+4)^6\left[9(x+x^{-1})^8\left(1-x^{-2}\right)\right] + (x+x^{-1})^9\left[6(3x+4)^5\left(3\right)\right]\)
\(\displaystyle =(3x+4)^5(x+x^{-1})^8\left((3x+4)\left[9\left(1-x^{-2}\right)\right] + (x+x^{-1})\left[6\left(3\right)\right]\right)\)
\(\displaystyle =(3x+4)^5(x+x^{-1})^8\left((27x+36)\left(1-x^{-2}\right) + 18(x+x^{-1})\right)\)
\(\displaystyle =(3x+4)^5(x+x^{-1})^8\left(27x+36-27x^{-1}-36x^{-2} + 18x+18x^{-1}\right)\)
\(\displaystyle =(3x+4)^5(x+x^{-1})^8\left(45x+36-9x^{-1}-36x^{-2}\right)\)
\(\displaystyle =9(3x+4)^5(x+x^{-1})^8\left(5x+4-\frac{1}{x}-\frac{4}{x^2}\right)\)

All done!
 
there must be a simple way than that

can you at least tell me the answer to the first problem, then i work backwards?
 
\(\displaystyle f(t) \ = \ (2t-1)^{3}(t^{2}+1)^{5}\)

\(\displaystyle f \ ' \ (t) \ = \ 3(2t-1)^{2}(2)(t^{2}+1)^{5}+(2t-1)^{3}(5)(t^{2}+1)^{4}(2t)\)

\(\displaystyle = \ 6(2t-1)^{2}(t^{2}+1)^{5}+10t(2t-1)^{3}(t^{2}+1)^{4}\)

\(\displaystyle = \ 2(2t-1)^{2}(t^{2}+1)^{4}[3(t^{2}+1)+5t(2t-1)]\)

\(\displaystyle = \ 2(2t-1)^{2}(t^{2}+1)^{4}(13t^{2}-5t+3)\)

\(\displaystyle Note: \ 13t^{2}-5t+3 \ is \ prime.\)
 
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