Results 1 to 7 of 7

Math Help - Help Pleas (d/dx)

  1. #1
    Newbie
    Joined
    Sep 2008
    Posts
    2

    Question Basic Calculas (d/dx)

    ok (d/dx) sin x= cos x

    therefore (d/dx)cos x= -sinx

    but how do u explain in terms how (d/dx)cosx = sin x

    i believe i am supposed to us the formula f'(x)= F(x + h) - f(x) / h

    but can someone show me how it is done so i can understand how it is expressed.
    Last edited by tir0nel; September 10th 2008 at 08:10 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Ok. So the limit definition by your notation should be something like f'(x)=\lim_{h \rightarrow 0} \frac{ \cos{(x+h)}-\cos{x}}{h}. So to handle the numerator use this identity - \cos{(a+b)}=\cos{(a)}\cos{(b)}-\sin{(a)}\sin{(b)}. Apply this to your limit and post back what you find.
    Last edited by Jameson; September 10th 2008 at 08:33 PM. Reason: messed up the limit a bit
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Thinking more that actually is a tricky limit. Try just using the fact that <br />
\sin{ \left( \frac{\pi}{2} + x \right)} = \cos{x} and that \sin(a+b) = \sin(a) \sin(b) + \cos(a) \cos(b)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    f'(x)=\lim_{h \rightarrow 0} \frac{ \cos{(x+h)}-\cos{x}}{h}

    \lim_{h \rightarrow 0}\frac{cos(x)cos(h)-sin(x)sin(h) - cos(x)}{h}

    \lim_{h \rightarrow 0}\frac{-sin(x)sin(h) -cos(x)(1- cos(h))}{h}

    -\lim_{h \rightarrow 0}\frac{sin(x)sin(h)}{h} -\lim_{h \rightarrow 0}\frac{cos(x)(1-cos(h))}{h}

    -sin(x)\lim_{h \rightarrow 0} \frac{sin(h)}{h}-cos(x)\lim_{h \rightarrow 0}\frac{1-cos(h)}{h}

    Then remember that

    \lim_{h \rightarrow 0} \frac{sin(h)}{h} = 1

    and

    \lim_{h \rightarrow 0} \frac{1-cos(h)}{h}=0

    You should be able to finish up from here and that was a tricky question lol
    Last edited by 11rdc11; September 10th 2008 at 09:06 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    ^ Well just do it all why don't you?

    Nicely done. OP that's exactly what you want.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    lol yep I was stuck for a good bit becuause the cos(x) in the numerator would cancel out leaving \frac{0}{0}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Yop !

    A more straighforward method is to use \cos(a)-\cos(b)=-2 \sin \frac{a+b}{2} \sin \frac{a-b}{2}

    \frac{\cos(x+h)-\cos(x)}{h}=-2 \cdot \frac{\sin(x+\tfrac h2) \sin(\tfrac h2)}{h}=-2 \cdot \sin(x+\tfrac h2) \cdot \frac{\sin(\tfrac h2)}{h}

    We know that \lim_{x \to 0} \frac{\sin(ax)}{x}=a. Therefore \lim_{h \to 0} \frac{\sin(\tfrac h2)}{h}=\frac 12

    And we have \lim_{h \to 0} \frac{\cos(x+h)-\cos(x)}{h}=-\sin(x)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: February 25th 2010, 02:15 AM
  2. pleas help. probability.
    Posted in the Statistics Forum
    Replies: 1
    Last Post: January 25th 2009, 10:20 PM
  3. [SOLVED] Pleas help me with PHYSICS!
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: November 30th 2007, 06:12 AM
  4. Conics Question Pleas Help****
    Posted in the Math Topics Forum
    Replies: 0
    Last Post: November 21st 2007, 04:52 PM
  5. Conic Problems help me pleas...
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: January 7th 2007, 09:33 AM

Search Tags


/mathhelpforum @mathhelpforum