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Math Help - Deivative by first principle.

  1. #1
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    Deivative by first principle.

    I am not able to find the derivative of the following function by first principle. I need the entire procedure of the first principle. Thanx in advance.

    f(x)= x^2.cosx
    Last edited by tariq_h_tauheed; September 12th 2008 at 12:08 PM.
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  2. #2
    MHF Contributor arbolis's Avatar
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    (u\times v)'(x)=u'(x)v(x)+u(x)v'(x).
    Put u(x)=x^2 so u'(x)=2x.
    Put v(x)=\cos (x) so v'(x)=-\sin (x). I let you finish this, it's just 2 multiplications and an addition.
    EDIT: Oops... sorry I didn't know what what the "first principle".
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  3. #3
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    You can differentiate using the chain rule.

    Use the identity e^(x^2)=2e^x

    so e^y=2e^(x.cosx)

    Differentiating,
    e^y(dy/dx)=2(cosx+(-xsinx))e^(x.cosx)

    Then just move e^y over and sub it as e^(x^2.cosx)
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by tariq_h_tauheed View Post
    I am not able to find the derivative of the following function by first principle. I need some urgent help. Thanx in advance.

    f(x)= x^2.cosx
    Guys, the question is from first principles, so we don't have access to the chain rule.

    f'(x) = \lim_{h \to 0} \frac{(x + h)^2~cos(x + h) - x^2~cos(x)}{h}

    Recall that cos(x + h) = cos(x)~cos(h) - sin(x)~sin(h), so expanding we get
    = \lim_{h \to 0} \frac{(x^2 + 2hx + h^2)~(cos(x)~cos(h) - sin(x)~sin(h)) - x^2~cos(x)}{h}

    = \lim_{h \to 0} \frac{(x^2 + 2hx + h^2)~cos(x)~cos(h) - (x^2 + 2hx + h^2) sin(x)~sin(h) - x^2~cos(x)}{h}

    = \lim_{h \to 0} \frac{(2hx + h^2)~cos(x)~cos(h) - (x^2 + 2hx + h^2) sin(x)~sin(h)}{h}

    = \lim_{h \to 0} \frac{(2hx + h^2)~cos(x)~cos(h)}{h} - \lim_{h \to 0}\frac{(x^2 + 2hx + h^2) sin(x)~sin(h)}{h}

    See if you can finish this.

    -Dan
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