Results 1 to 9 of 9

Math Help - Question about Derivative Rules

  1. #1
    Newbie
    Joined
    Nov 2009
    Posts
    15

    Question about Derivative Rules

    Hi, I just need clarification on the constant rule of derivatives.
    The derivative of a constant = 0. So if after applying the quotient or product rule I end up with,

    (x-4)d(2) + (2)d(x-4)
    =(x - 4)(0) + (2)(1)
    = 0 + 2

    Similarly,

    (x^2 + 2x + 1)d(2) + (2)d(x^2 + 2x + 1)
    = (x^2 + 2x + 1)(0) + (2)(2x + 2)
    = 0 + (4x + 4)

    (d = derivative, i don't know how else to enter it in here.)

    What I need clarification with is if we have a derivative of a constant multiplied by another function, does it always equal 0 because we multiply by 0? Or do we just remove the constant derivative?

    So if we just remove the constant derivative the first equation would end up like

    (x - 4) + (2x - 1) instead of 0 + (2x - 1).

    Thank you in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by abstraktz View Post
    (x-4)d(2) + (2)d(x-4)
    =(x - 4)(0) + (2)(1)
    = 0 + 2
    This is correct.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Calculus26's Avatar
    Joined
    Mar 2009
    From
    Florida
    Posts
    1,271
    I think you could save yourself some time if you realize simply

    d(cf)/dx = cdf/dx

    or for someting like 2/(x+1) d (2/(x+1))/dx = 2 d(1/(x+1))/dx
    Follow Math Help Forum on Facebook and Google+

  4. #4
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by abstraktz View Post
    Hi, I just need clarification on the constant rule of derivatives.
    The derivative of a constant = 0. So if after applying the quotient or product rule I end up with,

    (x-4)d(2) + (2)d(x-4)
    =(x - 4)(0) + (2)(1)
    = 0 + 2

    Similarly,

    (x^2 + 2x + 1)d(2) + (2)d(x^2 + 2x + 1)
    = (x^2 + 2x + 1)(0) + (2)(2x + 2)
    = 0 + (4x + 4)

    (d = derivative, i don't know how else to enter it in here.)

    What I need clarification with is if we have a derivative of a constant multiplied by another function, does it always equal 0 because we multiply by 0? Or do we just remove the constant derivative?

    So if we just remove the constant derivative the first equation would end up like

    (x - 4) + (2x - 1) instead of 0 + (2x - 1).

    Thank you in advance!
    If y=k \: \, \: k \neq f(x) then \frac{dy}{dx}(k) = 0.

    However, your question seems strange, what was the original question? You can represent derivatives with respect to x with [tex]\frac{d}{dx} f(x)[/tex]. In your case replace f(x) with the function to differentiate
    Follow Math Help Forum on Facebook and Google+

  5. #5
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    Quote Originally Posted by abstraktz View Post
    Hi, I just need clarification on the constant rule of derivatives.
    The derivative of a constant = 0. So if after applying the quotient or product rule I end up with,

    (x-4)d(2) + (2)d(x-4)
    =(x - 4)(0) + (2)(1)
    = 0 + 2

    Similarly,

    (x^2 + 2x + 1)d(2) + (2)d(x^2 + 2x + 1)
    = (x^2 + 2x + 1)(0) + (2)(2x + 2)
    = 0 + (4x + 4)

    (d = derivative, i don't know how else to enter it in here.)

    What I need clarification with is if we have a derivative of a constant multiplied by another function, does it always equal 0 because we multiply by 0? Or do we just remove the constant derivative?

    So if we just remove the constant derivative the first equation would end up like

    (x - 4) + (2x - 1) instead of 0 + (2x - 1).

    Thank you in advance!
    Note that \frac{d}{dx}[kf(x)]=k\frac{d}{dx}[f(x)] where k is a constant. What this means is that we can pull constants out of the differentiation process.

    So, in your first example:

    \frac{d}{dx}[2(x-4)]=2\frac{d}{dx}(x-4)=2(1)=2
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Nov 2009
    Posts
    15
    Thank you people!
    Um I made that question up on the spot earlier as I didn't have it with me. But the original question was

    \frac{d}{dx} 1/x-1
    Follow Math Help Forum on Facebook and Google+

  7. #7
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    Quote Originally Posted by abstraktz View Post
    Thank you people!
    Um I made that question up on the spot earlier as I didn't have it with me. But the original question was

    \frac{d}{dx} 1/x-1
    \frac{1}{x-1} or \frac{1}{x}-1?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Nov 2009
    Posts
    15
    <br />
\frac{1}{x-1}<br />

    which equals

    <br />
\frac{-1}{(x-1)^2}<br />
    Follow Math Help Forum on Facebook and Google+

  9. #9
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    Quote Originally Posted by abstraktz View Post
    <br />
\frac{1}{x-1}<br />

    which equals

    <br />
\frac{-1}{(x-1)^2}<br />
    Yes, \frac{d}{dx}\left[\frac{1}{x-1}\right]=\frac{-1}{(x-1)^2}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Derivative Rules..
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 22nd 2010, 02:35 PM
  2. Which derivative rules?
    Posted in the Calculus Forum
    Replies: 9
    Last Post: May 17th 2009, 12:31 PM
  3. A question related to Derivative Rules.
    Posted in the Calculus Forum
    Replies: 5
    Last Post: February 14th 2008, 04:03 PM
  4. Using Derivative rules...
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 11th 2008, 06:01 PM
  5. Backwards Derivative Rules
    Posted in the Calculus Forum
    Replies: 11
    Last Post: April 22nd 2007, 11:16 AM

Search Tags


/mathhelpforum @mathhelpforum