Expand (2x - 1) ^5
so if x^5 =x^3 *x^2 then how do you think ?
Use binomial expansion
\[=(2x)^{5}+\left(\begin{matrix}5 \\ 4\end{matrix}\right)(2x)^{4}(-1)+\left(\begin{matrix}5 \\ 3\end{matrix}\right)(2x)^{3}(-1)^{2}+\left(\begin{matrix}5 \\ 2\end{matrix}\right)(2x)^{2}(-1)^{3}+\left(\begin{matrix}5 \\ 1\end{matrix}\right)(2x)(-1)^{4}+(-1)^{5}\]
oops last term is (-1)^5 hope you get the idea...look at the link posted above as well
Thanks alot
(2x-1)^5 = ( 2 x - 1 ) ^ 5 i can expand more if you want... :)
Sure. Any help is welcomed hahaha
ok.. ( 2 x - 1 ) ^ 5
:)
Haha thanks for being so in-depth
too much math... :)
Get the coefficients from pascal's triangle and the terms from the binomial theorem. to find y and y', differentiat the function as is from the power rule. From the Pascal's Triangle, the coefficients are 1 5 10 10 5 1 http://en.wikipedia.org/wiki/Pascal's_tr … This site also has an explanation of how you can get these coeffiecients from the combination formula if Pascal's triangle is too unweildy for higher powers of binomial expansion From the binomial theorem the kth term is (2x)^(6-k)*(-1)^(k-1) So your expansion is 1(2x)^5(-1)^0 + 5(2x)^4(-1)1 + 10(2x)^3(-1)^2 + 10(2x)^2(-1)^3 + 5(2x)^1(-1)^4 + 1(2x)^0(-1)^5 Simplifying, this is 32x^5 - 80x^4 + 80x^3 - 40x^2 + 10x -1 (but don't take my word for it, work it out yourself to make sure) To find y' and y", you could take the derivative either by the power rule on (2x-1)^5 or on the expansion, whichever you prefer. Either one should give you the same result. I always prefer to work with given values rather than derived values, in case there are errors in the derivation. So rather than take the derivative of the expansion, I'll just use the power rule. y = (2x-1)^5 y' = (2 (5(2x-1)^4)) = 10(2x-1)^4 y" = (2(4(2x-1)^3)) = 8(2x-1)^3
Wow thank you so much. You made it seem easy lol
tats ok
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