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Thread: Derivation using the definition of a derivite

  1. #1
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    Derivation using the definition of a derivite

    The question I am struggling with is as follows:

    Derivation using the definition of a derivite-q2..jpg

    My working is as follows:
    Derivation using the definition of a derivite-q2w1.jpg

    Derivation using the definition of a derivite-q2w2.jpg
    Derivation using the definition of a derivite-q2w3.jpg
    Derivation using the definition of a derivite-q2w4.jpg
    Derivation using the definition of a derivite-q2w5.jpg
    I don't know how to continue from here or even rather if i'm on the right track.
    Thank you in advance for any help!
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  2. #2
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    Re: Derivation using the definition of a derivite

    Quote Originally Posted by ImPoob View Post
    The question I am struggling with is as follows:

    Click image for larger version. 

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    My working is as follows:
    Click image for larger version. 

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    Click image for larger version. 

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    I don't know how to continue from here or even rather if i'm on the right track.
    Thank you in advance for any help!
    You are making it entirely too difficult. First of all, you are asked to find $F'(0)$ so $x=0$ and you might better start with$$
    F'(0) = \lim_{h\to 0} \frac{F(0+h) - F(0)}{h} = \lim_{h\to 0} \frac {f(h)\sin^2(h)}{h}
    $$where I have omitted terms that are $0$. Now think about breaking that into the product of three factors, each of which is easy to find the limit.
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  3. #3
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    Re: Derivation using the definition of a derivite

    Thanks for you help! After mulling over your explanation I understand how you arrived at:

    Derivation using the definition of a derivite-fuq2.jpg
    When you say break it into a product of three factors I assume you simply mean:

    Derivation using the definition of a derivite-fuq22.jpg
    If this is correct I still wouldn't know how to continue from here, as in all the examples iv'e come across before the h in the denominator cancels. Currently we would be dividing by zero if we took the limit of the function. My only other thought was to rewrite sin^2 (x) in some other fashion using trig identities but that still doesn't help with the h in the denominator.

    Thanks a bunch again.
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  4. #4
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    Re: Derivation using the definition of a derivite

    No, that is NOT what he meant! For one thing you have mysteriously changed two of the "h"s to "x"s. Also you did not completely change it into a product of three terms.

    You should have $\displaystyle \left(\lim_{h\to 0} f(h)\right)\left(\lim_{h\to 0} sin(h)\right)\left(\lim_{h\to 0}\frac{sin(h)}{h}\right)$.
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  5. #5
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    Re: Derivation using the definition of a derivite

    Ah. Thank you for your help also!

    Would the final step be to use limit laws to evaluate the expression using lim h-> 0 sin(h)/h = 1?

    Thank again.
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  6. #6
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    Re: Derivation using the definition of a derivite

    Quote Originally Posted by ImPoob View Post
    Ah. Thank you for your help also!

    Would the final step be to use limit laws to evaluate the expression using $\displaystyle \mathop {\lim }\limits_{h \to 0} \frac{{\sin (h)}}{h} = 1$?
    Yes indeed!
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  7. #7
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    Re: Derivation using the definition of a derivite

    And what answer did you get for this problem?
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