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Differentiation Problem

  1. May 2, 2014 #1

    adjacent

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    1. The problem statement, all variables and given/known data
    Find the gradient of the tangent at x on the following curve
    ##y=3x^2##

    2. Relevant equations
    $$\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$


    3. The attempt at a solution
    I know that it's ##6x##.
    $$\frac{3(x+\Delta x)^2-3x^2}{\Delta x}$$
    $$=\frac{3x^2+6x\Delta x +3{\Delta x}^2-3x^2}{\Delta x}$$
    $$=\frac{3\Delta x(2x + \Delta x)}{\Delta x}=2x+3\Delta x$$
    $$\lim_{\Delta x\rightarrow 0}2x+3\Delta x =2x$$

    :confused:
    Where did I go wrong?
     
  2. jcsd
  3. May 2, 2014 #2

    Curious3141

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    In this step, have you distributed the 3 correctly on multiplying?
     
  4. May 2, 2014 #3

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    Ohhhh

    $$=\frac{3\Delta x(2x + \Delta x)}{\Delta x}$$
    $$=2\Delta x(2x+\Delta x)$$
    $$\lim_{\Delta x \to 0}2\Delta x(2x+\Delta x)=0$$

    Whaat?
     
    Last edited: May 2, 2014
  5. May 2, 2014 #4

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    I just realised that I made an algebra mistake.
    $$\frac{3\Delta x(2x + \Delta x)}{\Delta x}=3(2x+\Delta x)$$
    $$\lim_{\Delta x \to 0}6x+ \Delta x=6x$$
    :smile:

    How foolish I am!
     
    Last edited: May 2, 2014
  6. May 2, 2014 #5

    Curious3141

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    Believe me, I've been far more foolish on many occasions! :biggrin: Just a simple error, nothing to beat yourself up about.
     
  7. May 2, 2014 #6

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    Thanks.
    Another question came to mind,
    Differentiate ##f(x)=x^2+3x+1##
    I think the correct answer would be ##2x+3##

    But I have a doubt.
    I used the addition rule there , ##(v+u)'=v'+u'##
    So what do I do about the derivative of 1? wouldn't it be ∞?
     
  8. May 2, 2014 #7

    Curious3141

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    Why would it be infinite? The derivative is basically a rate of change. What is the rate of change of a constant function?
     
  9. May 2, 2014 #8

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    Oh. I thought it was x=1.

    Actually I am self-learning calculus.That's why I make these simple mistakes.

    Another question.

    ##(u+v)'=u'+v'##.
    So what are u and v here? Are they functions?
    In the function ##f(x)=x^2+3x+1##, is ##x^2##,##3x## and ##1## considered functions?Aren't they a part of the function ##f##?
     
  10. May 2, 2014 #9

    SteamKing

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    There's nothing that says f can't be the sum or product of other functions.
     
  11. May 2, 2014 #10

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    So ##x^2+3x+1## is treated like sum of functions?

    Q.
    Why is ##(\frac{1}{v})'=\frac{v'}{v^2}##? I saw it from here

    If I apply the power rule, I get
    ##v^{-1}=-v^{-1-1}=-v^{-2}=-\frac{1}{v^2}##

    Does that mean that rule is wrong?
     
  12. May 2, 2014 #11
    it said
    -f'/f^2
    so it is right, v' is just the derivative of the denominator and not the reciprocal, in your example, lets use that.
    v' = 1
    so (1/v)' = -1/v^2
     
  13. May 2, 2014 #12

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    ohh. So it's v itself!
    So what's the use of that rule? It can be easily derived from the power rule,right?

    EDIT:Using the quotient rule is easier.
    I just tried using the power rule for
    ##\frac{1}{2x^2+3x}## and it gave me ##-\frac{1}{(2x^2+3x)^2}## instead of ##-\frac{4x+3}{(2x^2+3x)^2}## which I got by using both the reciprocal and the quotient rule.
    :confused:
     
    Last edited: May 2, 2014
  14. May 2, 2014 #13
    Yea, I wouldn't suggest memorizing anything more than the chain rule, power rule, and product rule. That is all you need for differentiating, of course there is the quotient rule, but that can be derived with the product rule + chain rule + power rule. The sum and difference rules I never learned explicitly, they were just implied to be true.
     
  15. May 2, 2014 #14

    SteamKing

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    Yes. The derivative of the sum of several different functions is equal to the sum of the derivatives of each function. For example:

    g(x) = f1(x) + f2(x) + f3(x), where f1, f2, and f3 are all continuous functions of x.

    g'(x) = f1'(x) + f2'(x) + f3'(x), where the ' indicates the derivative w.r.t. x.

    This is also analogous to the limit of a sum being equal to the sum of the limits, so long as each limit exists.
     
  16. May 2, 2014 #15

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    What about those trigonometric things and exponents etc?

    Note: I have edited my last post there,please see it.
     
  17. May 2, 2014 #16
    Oh I meant rules, but yea the actual derivatives you will need to memorize, like (sin(x))' = cos(x) and (ln(x))' = 1/x, and for your edited post, so imagine
    v = (2x^2 + 3x)^(-1)
    then
    v' = -4x -3/(2x^2+3x)^(2)
    You get it now?
    Have you learned the chain rule yet? Usually I find that it is what most people have trouble with when learning differential calculus.
     
    Last edited: May 2, 2014
  18. May 3, 2014 #17

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    No.
    See what I did:
    $$\frac{d}{dx}(2x^2+3x)^{-1}=-1(2x^2+3x)^{-1-1}=-1(2x^2+3x)^{-2}$$
    $$=-\frac{1}{(2x^2+3x)^2}$$

    What is the wrong step here?

    I will learn it today.
     
  19. May 3, 2014 #18

    HallsofIvy

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    That is a difficulty with using the "Newton notation" for the derivative (primes) rather than the Leibniz notation- you don't make the variable explicit. It is true that [itex]dx^{-1}/dx=-x^{-2}[/itex]. And [itex]dy^{-1}/dy= -y^{-2}[/itex] and [itex]du^{-1}/du= -u^{-2}[/itex] and [itex]dv^{-1}/dv= -v^{-2}[/itex]. It doesn't matter what you call the variable!

    So if [itex]v= 2x^2+ 3x[/itex] then [itex](2x^2+ 3x)^{-1}= v^{-1}[/itex] so that
    [tex]\frac{d(2x^2+ 3x)^{-1}}{dv}= \frac{dv^{-1}}{dv}= -v^{-2}= -(2x^2+ 3x)^{-2}[/tex]

    But that is the derivative with respect to v (or with respect to [itex]2x^2+ 3x[/itex]), not with respect to x!
    To do that you can use the chain rule:
    [tex]\frac{d f(v)}{dx}= \left(\frac{df(v)}{dv}\right)\left(\frac{dv}{dx}\right)[/tex]
     
  20. May 3, 2014 #19

    Curious3141

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    You didn't apply chain rule here. You just differentiated treating ##2x^2+3x## as if it was x.

    Chain rule allows differentiation of composite functions. So if you have a composite function like ##h(x) = f(g(x))## and you want to find the derivative, it's given by ##h'(x) = f'(g(x)).g'(x)##. Note that the dot is often used to repent multiplication in unwieldy algebraic expressions.

    In your case ##f(x) = x^{-1}## while ##g(x) = 2x^2 + 3x##. Think about why this is so. You need to have a great grasp of functional theory before going on to calculus.

    The easiest albeit informal way I can describe the intuitive application of chain rule is as "treat function as a function of another part, then differentiate the function as if it were just a function of x BUT subbing the part in place of x and THEN multiply by the derivative of the part." This is how I think about Chain Rule when I actually apply it. You can iterate the rule as many time as you like and you'll see this when you read more and have to deal with nested functions like ##\sin^3(\cos(x))##.
     
  21. May 3, 2014 #20

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    So If I use the quotient formula (##(\frac{u}{v})'=\frac{u'v-uv'}{v^2}##), am I differentiation with respect to x?
    I don't think so.Why are we not applying chain rule in this case?
    Aren't we considering v as the denominator(##2x^2+ 3x##)?
    I have learned chain rule today.
    The chain rule is ##(fg(x))'=f'g(x) . g'(x)##
    How do I extend this to more than two functions?
    The lecture videos I watched did not explain chain rules well! But he said that if you know chain rule,you can conquer the world!(lol, I don't know why.Is this so useful?)
     
    Last edited: May 3, 2014
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