Question

find the differential of the dependent variable:

g(x)= (x^2 + 3x +1)((1/x) + (1/x^2) +3)

Answer 1

Fortunately, the "differential" is a relatively easy concept.
If y=g(x), then (dy/dx) = g '(x), and this can be re-written as dy = g '(x) dx.

So, in your math problem, to find the differential of the function g(x),

write dy = g '(x)*dx, after you've found the derivative of g '(x).

Answer 2

heeeeyy its you again! :D

Answer 3

If y = x^2, the derivative is dy/dx = 2x, and the differential is dy = 2xdx.

Re: "heeeeyy its you again!" Is that good? or...not? :)

Answer 4

good ofc

Answer 5

What a relief! :)

Answer 6

so i do the derivative of each parenthesis?

Answer 7

You're given the function g(x). Just find the derivative of g(x) in the usual way. Notice that your g(x) is a PRODUCT, so use the product rule.

Answer 8

im not sure what the derivative of the fractions are

Answer 9

The second multiplicand is \[((1/x) + (1/x^2) +3)\]

Answer 10

...and it's probably best to rewrite this before taking the derivative:\[x ^{-1}+x ^{-2}+3\]

Answer 11

From this point on, you just apply the power rule to each term.

For example, the deriv. of the first term is -x^(-2). What are the derivs. of x^(-2) and 3?

Answer 12

im sorry where did you get the -x^(-2) from?

Answer 13

\[\frac{ d }{ dx }(\frac{ 1 }{ x })=\frac{ d }{ dx }x ^{-1}=-1x ^{-2}\]by the Power Rule.

Answer 14

ah yes i forgot

Answer 15

so itd be (2x+3) + (-x^-2 - x^-3)

Answer 16

So, \[\frac{ d }{ dx }(\frac{ 1 }{ x }+\frac{ 1 }{ x^2 }+3) = -1x ^{-2} -2x ^{-3}+0\]

Answer 17

ops forgot the 2

Answer 18

Rememb\[\frac{ d }{ dx }u*v=u \frac{ dv }{ dx }+v \frac{ du }{ dx }\]er that the Product will crease a result with two terms and four factors:

Answer 19

create (not crease)

Answer 20

so that the derivative of your function g(x) will come out to (x^2+3x+1)(-x^(-2) -2x^(-3)) + ( )( ), two terms and four factors.

Answer 21

\[\frac{ d }{ dx }=(x^2+3x+1)(-x ^{-2}-2x ^{-3}) + (~)*(~)\]

Answer 22

yup got that

Answer 23

can you fill in the two empty sets of parentheses?


Good.

so, supposing you end up with g '(x). Your differential will then be\[dy=dg=g '(x)*dx.\]

Answer 24

so its dg(x) = (x^2 + 3x +1)(-x^-2 - 2x^-3) + (1/x + 1/x^2 +3) (2x+3) dx

Answer 25

\[dg = [(x^2 + 3x +1)(-x^-2 - 2x^-3) + (1/x + 1/x^2 +3) (2x+3)] dx\]

Answer 26

the only significant change I've made here was to insert brackets: [ ].

Answer 27

thanks! how long will you be on OS for because i have quite a few questions and i was wondering if you may help me with them

Answer 28

Not sure, really. I've spent a significant part of my day on OS, and realize I'm overdoing it. Nevertheless: go ahead and post one of your more challenging remaining questions.