Question

Lex f(x) =5x ^{6}sqrt{x}+6/(x ^{3}sqrt{x}).
Find f'(x)

Answer 1

is this what you mean

\[5\,{x}^{13/2}+6\,{x}^{-7/2}\]

Answer 2

\[5x ^{6}\sqrt{x}+6/(x ^{3}\sqrt{x})\]

Answer 3

lol same thing :)

Answer 4

whoops, then yes lol

Answer 5

well use the same rule i showed you earlier plus this rule

f'(x)=a'(x)+b'(x) something like that meaning you can take the derivitive of each term individually.

Answer 6

So then is it \[65/2x ^{11/2}-21x ^{-9/2}\]

Answer 7

x^(13/2)+6/x^(7/2)

13/2 x^2(13/2-1) + -7/2*6^(-7/2-1)

Answer 8

after that just multiply out whatever needs to be multiplied the (-7/2) * 6

subtract the ones inside the exponents to make the fractions simplified

then
result is (65/2)*x^(11/2)-21/x^(9/2)

\[{\frac {65}{2}}\,{x}^{11/2}-21\,{x}^{-9/2}\]

Answer 9

You are so right :D wow you dont need me at all lol.

Answer 10

Haha thanks! so what is f'(6)?

Answer 11

That answer above was wrong

Answer 12

plug in 6 at all the xs

(655050233/2592)*sqrt(6)

Answer 13

which is 619035.041181

Answer 14

they're both incorrect

Answer 15

:(

Answer 16

I'm not sure why.

Answer 17

i see i think it is because of the 21

Answer 18

how odd i redid all and still get same :(

Answer 19

and i did it with my calculator and it got me same result manually

Answer 20

hmmm....so is the answer still the same?

Answer 21

you get 619035.041181 from the 1st term. Did you subtract the 2nd?

Answer 22

I typed that in as well and those aren't correct.

Answer 23

http://www.wolframalpha.com/input/?i=%2865%2F2%29*x%5E%2811%2F2%29-21%2Fx%5E%289%2F2%29%2C+x%3D6

Answer 24

Mb limit for decimals or asked for f(6), not f'(6)?

Answer 25

f'(x)=2/3[x^(15/2)]-12/5[x^(-5/2)] + c......(X) (c is a constant)




put x=6 and get c the resubstitute in eq. X and get f'(x)......