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Math Help - Using formal definition to find derivative

  1. #1
    Member zzizi's Avatar
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    Using formal definition to find derivative

    Hi

    I have to find the derivative of the following function: xsinx
    Using formal definition. I have tried to start this but am stuck. Can someone give me a hint to help me complete it?

    What I have done so far..

    lim as h approaches 0

    (x+h)(sinx+h) - (xsinx)/h <---- is this correct? or have I missed something?
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  2. #2
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    Re: Using formal definition to find derivative

    Strictly speaking, what you have is NOT correct but only because you are missing parentheses. The derivative is given by the limit of ((x+h)sinx+h- xsinx)/h. What you wrote would be (x+y)sinx+h- (xsinx/h). Even clearer would be ((x+h)sin(x+h)- xsin(x))/h. Now I would recommend that you use "-a+ a" to isolate the changes:
    ((x+h)sin(x+h)- xsin(x+h)+ xsin(x+h)- xsin(x))/h and separate into (((x+h)- x)/h)sin(x)+ x((sin(x+h)- sin(x))/h.
    The first you should recognise as the derivative of x and the second as the derivative of sine.
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  3. #3
    Member zzizi's Avatar
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    Re: Using formal definition to find derivative

    Thank you very much
    Last edited by zzizi; November 24th 2012 at 01:20 PM.
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    Re: Using formal definition to find derivative

    using the definition:

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

    for f(x) = xsin(x), we get:

    f'(x) = \lim_{h \to 0} \frac{(x+h)\sin(x+h) - x\sin(x)}{h}

    = \lim_{h \to 0}\left[\frac{x\sin(x+h)- x\sin(x)}{h} + \frac{h\sin(x+h)}{h} \right]

    can you continue?
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  5. #5
    Member zzizi's Avatar
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    Re: Using formal definition to find derivative

    Thank you Deveno!!!

    Yes I can
    Last edited by zzizi; November 24th 2012 at 01:27 PM.
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  6. #6
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    Re: Using formal definition to find derivative

    Hello, zzizi!

    I just had to try this myself . . .

    We need two theorems:

    . . \lim_{h\to0}\frac{\sin h}{h} \:=\:1 \qquad \lim_{h\to0}\frac{\cos h - 1}{h} \:=\:0


    Using the formal definition, differentiate: f(x) \,=\,x\sin x

    f(x+h) - f(x) \;=\;(x+h)\sin(x+h) - x\sin x

    . . . . . . . . . . . . =\;x\sin(x+h) + h\sin(x+h) - x\sin x

    . . . . . . . . . . . . =\;x\big[\sin(x+h) - \sin x\big] + h\sin(x+h)

    . . . . . . . . . . . . =\;x(\sin x\cos h + \cos x\sin h - \sin x) + h(\sin x\cos h + \cos x\sin h)

    . . . . . . . . . . . . =\;x\big[\sin x(\cos h -1) + \cos x\sin h\big] + h\left(\sin x\cos h + \cos x\sin h\right)


    \frac{f(x+h)-f(x)}{h} \:=\:x\left[\sin x\!\cdot\!\frac{\cos h - 1}{h} + \cos x\!\cdot\!\frac{\sin h}{h}\right] +\sin x\cos h+\cos x\sin h


    \lim_{h\to0}\frac{f(x+h)-f(x)}{h} \;=\;x\left[\sin x\left(\lim_{h\to0}\frac{\cos h-1}{h}\right) + \cos x\left(\lim_{h\to0}\frac{\sin h}{h}\right)\right]

    . . . . . . . . . . . . . . . . . . . + \sin x\left(\lim_{h\to0}\cos h\right) + \cos x\left(\lim_{h\to0}\sin h\right)


    Therefore: . f'(x) \;=\;x\big[\sin x\cdot 0 + \cos x \cdot 1\big] + \sin x\cdot 1 + \cos x\cdot 0

    . . . . . . . . . f'(x) \;=\;x\cos x + \sin x . . ta-DAA!


    Edit: Ah, too slow again!
    . . . .Oh well, I simply refuse to delete all this typing.
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  7. #7
    Member zzizi's Avatar
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    Re: Using formal definition to find derivative

    Quote Originally Posted by Soroban View Post
    Hello, zzizi!

    I just had to try this myself . . .

    We need two theorems:

    . . \lim_{h\to0}\frac{\sin h}{h} \:=\:1 \qquad \lim_{h\to0}\frac{\cos h - 1}{h} \:=\:0



    f(x+h) - f(x) \;=\;(x+h)\sin(x+h) - x\sin x

    . . . . . . . . . . . . =\;x\sin(x+h) + h\sin(x+h) - x\sin x

    . . . . . . . . . . . . =\;x\big[\sin(x+h) - \sin x\big] + h\sin(x+h)

    . . . . . . . . . . . . =\;x(\sin x\cos h + \cos x\sin h - \sin x) + h(\sin x\cos h + \cos x\sin h)

    . . . . . . . . . . . . =\;x\big[\sin x(\cos h -1) + \cos x\sin h\big] + h\left(\sin x\cos h + \cos x\sin h\right)


    \frac{f(x+h)-f(x)}{h} \:=\:x\left[\sin x\!\cdot\!\frac{\cos h - 1}{h} + \cos x\!\cdot\!\frac{\sin h}{h}\right] +\sin x\cos h+\cos x\sin h


    \lim_{h\to0}\frac{f(x+h)-f(x)}{h} \;=\;x\left[\sin x\left(\lim_{h\to0}\frac{\cos h-1}{h}\right) + \cos x\left(\lim_{h\to0}\frac{\sin h}{h}\right)\right]

    . . . . . . . . . . . . . . . . . . . + \sin x\left(\lim_{h\to0}\cos h\right) + \cos x\left(\lim_{h\to0}\sin h\right)


    Therefore: . f'(x) \;=\;x\big[\sin x\cdot 0 + \cos x \cdot 1\big] + \sin x\cdot 1 + \cos x\cdot 0

    . . . . . . . . . f'(x) \;=\;x\cos x + \sin x . . ta-DAA!


    Edit: Ah, too slow again!
    . . . .Oh well, I simply refuse to delete all this typing.
    That's wonderful Soroban! But why was it slow? what happened?
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  8. #8
    MHF Contributor MarkFL's Avatar
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    Re: Using formal definition to find derivative

    Apparently it took soroban over two hours to type his response...
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  9. #9
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    Re: Using formal definition to find derivative

    Quote Originally Posted by MarkFL2 View Post
    Apparently it took soroban over two hours to type his response...
    All those god damn brackets and parenthesis.
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  10. #10
    MHF Contributor MarkFL's Avatar
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    Re: Using formal definition to find derivative

    He is the man when it comes to \LaTeX!!!
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  11. #11
    Member zzizi's Avatar
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    Re: Using formal definition to find derivative

    Oh right! Thank you very much Soroban for the effort, it has helped me understand
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  12. #12
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    Re: Using formal definition to find derivative

    If I was to do this using first principles, I'd evaluate the derivatives of \displaystyle \begin{align*} x \end{align*} and \displaystyle \begin{align*} \sin{x} \end{align*} by first principles, then prove the product rule using first principles and apply it

    \displaystyle \begin{align*} f_1(x) &= x \\ f_1 ' (x) &= \lim_{h \to 0}\frac{f_1(x + h) - f_1(x)}{h} \\ &= \lim_{h \to 0}\frac{x + h - x}{h} \\ &= \lim_{h \to 0}\frac{h}{h} \\ &= \lim_{h \to 0} 1 \\ &= 1 \\ \\ f_2(x) &= \sin{(x)} \\ f_2 '(x) &= \lim_{h \to 0}\frac{f_2(x + h) - f_2(x)}{h} \\ &= \lim_{h \to 0}\frac{\sin{(x+h)} - \sin{(x)}}{h} \\ &= \lim_{h \to 0}\frac{\sin{(x)}\cos{(h)} + \cos{(x)}\sin{(h)} - \sin{(x)}}{h} \\ &= \sin{(x)}\lim_{h \to 0}\frac{\cos{(h)} - 1}{h} + \cos{(x)}\lim_{h \to 0}\frac{\sin{(h)}}{h} \\ &= \sin{(x)} \cdot 0 + \cos{(x)} \cdot 1 \\ &= \cos{(x)} \end{align*}

    Now as for the product rule...

    \displaystyle \begin{align*} \left[ u(x) v(x) \right]' &= \lim_{h \to 0}\frac{u(x + h)v(x + h) - u(x)v(x)}{h} \\ &= \lim_{h \to 0}\frac{u(x+h)v(x+h) - u(x +h)v(x) + u(x +h)v(x) - u(x)v(x)}{h} \\ &= \lim_{h \to 0}u(x+h) \cdot \lim_{h \to 0}\frac{v(x+h) - v(x)}{h} + v(x)\cdot \lim_{h \to 0}\frac{u(x+h) - u(x)}{h} \\ &= u(x) v'(x) + u'(x) v(x)  \end{align*}

    So that means

    \displaystyle \begin{align*} \left[ x\sin{(x)} \right]' &= x \left[ \sin{(x)} \right]' + (x)' \sin{(x)} \\ &= x\cos{(x)} + \sin{(x)} \end{align*}
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  13. #13
    Member zzizi's Avatar
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    Re: Using formal definition to find derivative

    Quote Originally Posted by Prove It View Post
    If I was to do this using first principles, I'd evaluate the derivatives of \displaystyle \begin{align*} x \end{align*} and \displaystyle \begin{align*} \sin{x} \end{align*} by first principles, then prove the product rule using first principles and apply it

    \displaystyle \begin{align*} f_1(x) &= x \\ f_1 ' (x) &= \lim_{h \to 0}\frac{f_1(x + h) - f_1(x)}{h} \\ &= \lim_{h \to 0}\frac{x + h - x}{h} \\ &= \lim_{h \to 0}\frac{h}{h} \\ &= \lim_{h \to 0} 1 \\ &= 1 \\ \\ f_2(x) &= \sin{(x)} \\ f_2 '(x) &= \lim_{h \to 0}\frac{f_2(x + h) - f_2(x)}{h} \\ &= \lim_{h \to 0}\frac{\sin{(x+h)} - \sin{(x)}}{h} \\ &= \lim_{h \to 0}\frac{\sin{(x)}\cos{(h)} + \cos{(x)}\sin{(h)} - \sin{(x)}}{h} \\ &= \sin{(x)}\lim_{h \to 0}\frac{\cos{(h)} - 1}{h} + \cos{(x)}\lim_{h \to 0}\frac{\sin{(h)}}{h} \\ &= \sin{(x)} \cdot 0 + \cos{(x)} \cdot 1 \\ &= \cos{(x)} \end{align*}

    Now as for the product rule...

    \displaystyle \begin{align*} \left[ u(x) v(x) \right]' &= \lim_{h \to 0}\frac{u(x + h)v(x + h) - u(x)v(x)}{h} \\ &= \lim_{h \to 0}\frac{u(x+h)v(x+h) - u(x +h)v(x) + u(x +h)v(x) - u(x)v(x)}{h} \\ &= \lim_{h \to 0}u(x+h) \cdot \lim_{h \to 0}\frac{v(x+h) - v(x)}{h} + v(x)\cdot \lim_{h \to 0}\frac{u(x+h) - u(x)}{h} \\ &= u(x) v'(x) + u'(x) v(x)  \end{align*}

    So that means

    \displaystyle \begin{align*} \left[ x\sin{(x)} \right]' &= x \left[ \sin{(x)} \right]' + (x)' \sin{(x)} \\ &= x\cos{(x)} + \sin{(x)} \end{align*}
    Thank you very much for your input Prove it - much appreciated!
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