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Math Help - Use Definition to find Derivative

  1. #1
    Member VitaX's Avatar
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    Use Definition to find Derivative

    Find the derivative of Secx using the definition below:

    \frac{d}{dx}(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}

    Note y=Secx \rightarrow Cos^{-1}x

    \frac{d}{dx}(Secx) = \lim_{h \rightarrow 0} \frac{Cos^{-1}(x + h) - Cos^{-1}(x)}{h}

    = \lim_{h \rightarrow 0} \frac{1}{h}\left(\frac{1}{Cos(x + h)} - \frac{1}{Cos(x)}\right)

    = \lim_{h \rightarrow 0} \frac{1}{h}\left(\frac{Cos(x) - Cos(x + h)}{Cos(x + h)Cos(x)}\right)

    = \lim_{h \rightarrow 0} \frac{1}{h}\left(\frac{Cos(x) - (CosxCosh - SinxSinh)}{(CosxCosh - SinxSinh)Cos(x)}\right)

    I used the Identity Cos(x + h) = CosxCosh - SinxSinh

    After this I'm confused on where to go next. And this is supposed to be proving that \frac{d}{dx}(Secx)=SecxTanx I'm able to find this with the chain rule but been a while since I used the definition
    Last edited by VitaX; October 14th 2009 at 03:14 PM.
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  2. #2
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    Quote Originally Posted by VitaX View Post
    Find the derivative of Secx using the definition below:

    f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}
    Note y=Secx \rightarrow Cos^{-1}x

    f'(Secx) = \lim_{h \rightarrow 0} \frac{Cos^{-1}(x + h) - Cos^{-1}(x)}{h}
    f'(Secx) = \lim_{h \rightarrow 0} \frac{1}{h}\left(\frac{1}{Cos(x + h)} - \frac{1}{Cos(x)}\right)
    f'(Secx) = \lim_{h \rightarrow 0} \frac{1}{h}\left(\frac{Cos(x) - Cos(x + h)}{Cos(x + h)Cos(x)}\right)

    After this I'm confused on where to go next. And this is supposed to be proving that \frac{d}{dx}(Secx)=SecxTanx I'm able to find this with the chain rule but been a while since I used the definition
    first off , \cos^{-1}{x} is the notation for the arccosine (inverse cosine) function, not the reciprocal function \sec{x}.

    here is the correct notation ...

    \sec{x} = (\cos{x})^{-1}



    \lim_{h \to 0} \frac{1}{h} \left(\frac{\cos{x} - \cos(x+h)}{\cos(x+h)\cos{x}}\right)

    \lim_{h \to 0} \frac{1}{h} \left(\frac{\cos{x} - \cos{x}\cos{h}+\sin{x}\sin{h}}{\cos(x+h)\cos{x}}\r  ight)

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

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

    \lim_{h \to 0} \frac{1-\cos{h}}{h} \cdot \frac{\cos{x}}{\cos(x+h)\cos{x}}+ \frac{\sin{h}}{h} \cdot \frac{\sin{x}}{\cos(x+h)\cos{x}}

    recall the two basic trig limits ...

    \lim_{t \to 0} \frac{1 - \cos{t}}{t} = 0 and \lim_{t \to 0} \frac{\sin{t}}{t} = 1

    \lim_{h \to 0} \,\, 0 \cdot \frac{\cos{x}}{\cos(x+h)\cos{x}}+ 1 \cdot \frac{\sin{x}}{\cos(x+h)\cos{x}} = \frac{\sin{x}}{\cos^2{x}} = \frac{1}{\cos{x}} \cdot \frac{\sin{x}}{\cos{x}} = \sec{x} \tan{x}
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  3. #3
    Member VitaX's Avatar
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    Cos^{-1}x is the same as (Cosx)^{-1} Atleast that is what my professor always stresses because some people get that confused with Cosx^{-1} which is just the x value to that power. Thanks for the help but how come the bottom is not changed into the trig identity as well?

    Quote Originally Posted by skeeter View Post
    first off , \cos^{-1}{x} is the notation for the arccosine (inverse cosine) function, not the reciprocal function \sec{x}.

    here is the correct notation ...

    \sec{x} = (\cos{x})^{-1}



    \lim_{h \to 0} \frac{1}{h} \left(\frac{\cos{x} - \cos(x+h)}{\cos(x+h)\cos{x}}\right)

    \lim_{h \to 0} \frac{1}{h} \left(\frac{\cos{x} - \cos{x}\cos{h}+\sin{x}\sin{h}}{\cos(x+h)\cos{x}}\r  ight)

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

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

    \lim_{h \to 0} \frac{1-\cos{h}}{h} \cdot \frac{\cos{x}}{\cos(x+h)\cos{x}}+ \frac{\sin{h}}{h} \cdot \frac{\sin{x}}{\cos(x+h)\cos{x}}

    recall the two basic trig limits ...

    \lim_{t \to 0} \frac{1 - \cos{t}}{t} = 0 and \lim_{t \to 0} \frac{\sin{t}}{t} = 1

    \lim_{h \to 0} \,\, 0 \cdot \frac{\cos{x}}{\cos(x+h)\cos{x}}+ 1 \cdot \frac{\sin{x}}{\cos(x+h)\cos{x}} = \frac{\sin{x}}{\cos^2{x}} = \frac{1}{\cos{x}} \cdot \frac{\sin{x}}{\cos{x}} = \sec{x} \tan{x}
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  4. #4
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    you're welcome.
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  5. #5
    Member VitaX's Avatar
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    Thank you
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