Question

differentiate f(x)= 1/2x^6+3x^4-3/4x^3+1/3x^2-7/x+x-4

Answer 1

hehe now this one looks fun.

Answer 2

its hell

Answer 3

break it up into pieces using the sum rule

Answer 4

so lets do this, shall we ?

Answer 5

yes sir

Answer 6

so looking at : http://www.allaboutcircuits.com/vol_5/chpt_6/6.html

we want to break it up using the sum rule, so how would that look to you ?

Answer 7

and then once you have that broken up.. you break it up even further :)

Answer 8

o0

Answer 9

you got it ?

Answer 10

3x^5+12x^3-21/4x^2+

Answer 11

then the x in the denominator is like 0o

Answer 12

is it fine so far?

Answer 13

hmm let me see

Answer 14

taking the derivitive of the whole equation

Answer 15

derivative*

Answer 16

1/2x^6+3x^4-3/4x^3+1/3x^2-7/x+x-4

Answer 17

so we start wtih this now if we break it up ..

Answer 18

dx(1/2^6) + dx(3x^4) - dx(3/4x^3) + dx(1/3x^2) - dx(7/x) +dx(x) - dx(4)

Answer 19

what is the derivative of a constant? 0 because it never changes so we can eliminate that

Answer 20

yea

Answer 21

so simplify the simple ones first

Answer 22

1/2x^6 = 3x^5

Answer 23

we know the dx(x) = x

Answer 24

so now we have:

dx(1/2^6) + dx(3x^4) - dx(3/4x^3) + dx(1/3x^2) - dx(7/x) + x

Answer 25

3x^5+12x^3-2(1/4)x^2

Answer 26

left side follows this order:

power rule, power rule, power rule, power rule, now sure of dx(7/x)

Answer 27

then the denominator with 3x^2

Answer 28

just leave it was 1/6x?

Answer 29

notsure of dx(7/x)

Answer 30

would that just be 7?

Answer 31

-7x^2

Answer 32

I hope this helps?

Answer 33

yea, put me in the right direction, thanks

Answer 34

oh wait dx (7/x ) uses the quotient rule

using the quotient rule:

f(x) = 7
g(x) = x

x * dx(7) - 7 * dx(x) / (x^2)

Answer 35

which simplfies into -7x / x^2

Answer 36

ooh

Answer 37

thanks man big help

Answer 38

great to hear.

Answer 39

\[\frac{x^6}{2}+3 x^4-\frac{3 x^3}{4}+\frac{x^2}{3}+x-\frac{7}{x}-4 \]The derivative of the above is:\[3 x^5+12 x^3-\frac{9 x^2}{4}+\frac{7}{x^2}+\frac{2 x}{3}+1 \]