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Math Help - Derivatives - Tangent Lines

  1. #1
    Member purplec16's Avatar
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    Derivatives - Tangent Lines

    Ok so you have to use the first principle of derivates

    f(x)=5

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

    \lim_{h\to \0}\frac{\5+h-5}{h}

    After this step everything cancels out and the derivative is equal to zero....but I don't think it's suppose to be equal to zero
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  2. #2
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    Why do you don't think that? What the rate of change of a horizontal line? If you move one unit to the right, how much does it go up?
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  3. #3
    Member purplec16's Avatar
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    I guess because we always solved before for the limit not to be equal to zero.

    If you move one unit to the right it doesn't go up...i.e. it goes up by zero
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  4. #4
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    There you have it!

    Is your expectation unchanged or are you buying it?
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  5. #5
    Member purplec16's Avatar
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    both...so your saying that algebra (i.e. by smiplification) does not come into place to make the limit not equal to zero.
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    Hmmm... Not clear enough.

    The rate of change of a horizontal line is zero. If you work through the definition (without errors) and get a result that is different from zero, something is very wrong with the world.

    You would not be the first with an expectation that a solid definition corrected.
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  7. #7
    Member purplec16's Avatar
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    Okay thank you so much I guess I'll just stick with the facts then
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  8. #8
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    Quote Originally Posted by purplec16 View Post
    Ok so you have to use the first principle of derivates

    f(x)=5

    \lim_{h\to \0}\frac{\f(x+h)-f(x)}{h}
    Your "first principle" looks wrong. You have a "\" before the f and LaTeX did not recognize "\f" so left it all out.

    You meant \lim_{h\to 0}\frac{f(x+h)- f(x)}{h}

    \lim_{h\to \0}\frac{\5+h-5}{h}
    However, you have either propagated that mistake or do not know how to calculate f(x+h). With f(x)= 5, a constant, then f(x+ h)= 5.

    After this step everything cancels out and the derivative is equal to zero....but I don't think it's suppose to be equal to zero
    Now you have me really confused! Your incorrect \lim_{h\to\0} \frac{5+h- 5}{h}= \lim_{h\to \0}\frac{h}{h}= 1, not 0!

    What you should have had \lim_{x\to \0} \frac{5- 5}{h}= \lim_{x\to 0} \frac{0}{h}= 0 does give 0.

    However, I would consider the real "first principle" of the derivative to be "the derivative of a function, at a point, measures how fast that function is increasing at that point". A constant function is not changing at all- its "rate of change" and so it derivative, is 0.

    Even more fundamental: the derivative of a linear function, f(x)= mx+ b, is its slope, m. Of course, a constant function, f(x)= b, has m= 0.
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  9. #9
    Member purplec16's Avatar
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    yes i got that in the end thank you
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