What is the Answer to these Math Problem?

I am needing a little help with a few of these math problems.

Find the difference quotient to this problem:

1
__
2x


Please let me know. Thanks!

 

its not one half x, but 1 over 2x

 

Both are the same, so WHAT IS YOUR QUESTION???

 

You need to put the other end of the equation to ask for an answer.

 

this is a nonsense

 

this isn't a complete problem
or did you mean differential coefficient instead of difference quotient?

 

Hi,

I think this is not a complete question and you should write here the complete question. Please let us understand clearly.

Stella

 

I think you mean, what is the equation?

Try equating it with 1 OR 0.
You get answers as 1/2 and incase of 0, x gets eliminated.

 

in case of 0, it doesn't get eliminated, the answer comes out to be ∞ if x=0
but they're not the solutions, the question provides insufficient information for us to be able to solve it
Lets just wait for the OP to come online shall we?

 

Question make no sense, so answer will be nill from my side.

 

Korrupt I hate to tell you but your wrong, if x=0 mathematically it is undefined. Anything divided by 0 is undefined. Unless this is a calculus problem where you could argue that the limit of the function will go to infinity. But its not calculus so its just junk anyway.

 

I looked at your previous posts, I think you want the derivative of 1/2x using the quotient rule right? Use it as a product rule instead of (1/2)*x^-1 The 1/2 stays constant, you pull the -1 in front and subtract 1 from the exponent. You get -1/2x^-2 or -1/2x^2.

By quotient rule it would be Lower Function (2x) X Derivative High (0) - Higher (1) X Lower (2) / lower squared (4x^2) Same answer.

 

If you ever need tutoring let me know!

 

what the answer may i know?

 

this is really confusing...what exactlyy is the question everyone is givin their own solutions acoording to their interpretation of the question..lol

 

Let's see now....mmmmm.......so:

if we set 1 | x2

u = ex + e-x
then we find
du = (ex - e-x) dx

substitute du= (ex - e-x) dx, u = ex + e-x

solve

= ln |u| + C

substitute back u = ex + e-x

= ln |ex + e-x| + C

since ex and e-x are always positive

= ln (ex + e-x) + C

since (ex + e-x)/2 = cosh(x)

= ln (2 cosh x) + C
= ln 2 + ln (cosh x) + C

ln 2 is merely a constant that can be combined with C

= ln (cosh x) + 2

Your answer = +2

.jik

 

d/dx(1/2x) = 1/2(x^2) , hope you like my attempt to solve your homework

 

what we did was anything /0 is infinity, and 0/0 is undefined since it can have 3 answers:
1. 0/0=1, since anything divided by itself gives 1
2. 0/0=0, as 0 divided by anything =0
3. 0/0=infinity, as anything divided by 0 gives infinity

We are in the general section

 

1
__ =0
2x

=>
1=0

Try it out on paper..

 

Both aren't equal.....