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Urgent help please -- Differentiation technique question

  1. May 17, 2016 #1
    • Poster has been reminded to post schoolwork in the Homework Help forums
    I'm revising differentiation for tomorrow, I am hopelessly stuck on some things.

    I have some examples with answers, but I have no idea which rules are being used.

    y=2sin3x dy/dx=6cos3x

    I see that sin goes to cos but why is the two multiplied by the 3 (if that's what's happening)???

    y=3sin1/2x+e^x dy/dx= 6cos1/2x+e^x

    Thank you
     
  2. jcsd
  3. May 17, 2016 #2

    fresh_42

    Staff: Mentor

    It is the chain rule and in your second equation there might be something wrong.
     
  4. May 17, 2016 #3

    pasmith

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    [tex]\frac{d}{dx} \sin kx = k \cos kx[/tex] for constant [itex]k[/itex], as can be established by application of the chain rule.

    I wish you luck, but I fear you have left your revision too late.
     
  5. May 17, 2016 #4
    Okay thank you for the help, I have one more question which seems to need the chain rule, although I have an answer for it which looks false

    Here is the question and the answer I was given:

    y=3sin1/2x dy/dy=6cos1/2x

    And the answer I got with chain rule:

    1.5cos1/2x
     
  6. May 17, 2016 #5

    Mark44

    Staff: Mentor

    Your answer is correct, if I can interpret what you think you are writing.

    Please use parentheses!!
    Without parentheses, any of these are possible interpretations of the cosine part.
    ##\frac{\cos(1)}{2x}##
    ##\frac{\cos(1)}{2}x##
    ##\cos(\frac 1 2) x##
    ##\cos(\frac 1 {2x})##

    I presume you meant ##\frac 3 2 \cos(\frac x 2)##. As inline text, this should be written as (3/2) cos(x/2).
     
  7. May 17, 2016 #6
    I'm moving on to integration now. Here is my attempt at the first question:

    f(x)=sin 3x dx

    y=-cos 3x^2/2
     
  8. May 17, 2016 #7

    fresh_42

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    Control the equation Check your answer by differentiation. Where does the square come from?
     
    Last edited by a moderator: May 17, 2016
  9. May 17, 2016 #8
    I increased the power of 3x by 1.
     
  10. May 17, 2016 #9

    fresh_42

    Staff: Mentor

    Yes, I saw, but why? You want to integrate the sine function, not x.
     
  11. May 17, 2016 #10
    Isn't that hoe you integrate a function? For example 2x---> 2x^2/2
     
  12. May 17, 2016 #11

    fresh_42

    Staff: Mentor

    Yes. Here you have a linear function ##x → 2x## to integrate. But the function you want to integrate is ##x → \sin 3x##. You are right that it will give you something with ##- \cos##. But what would the derivation of ##- \cos \frac{3}{2}x^2## be?
     
  13. May 17, 2016 #12
    3/2sin^-1*3/2x^2
     
  14. May 17, 2016 #13

    SteamKing

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    That's not right. It's not even wrong.
     
  15. May 17, 2016 #14
    I have no idea how to do it, I know sin goes to -cos, and I thought 3x went to 3x^2/2, that's as far as I got using the maths is fun website.
     
  16. May 17, 2016 #15

    fresh_42

    Staff: Mentor

    Almost. The chain rule goes: ##(f \circ g)'(x) = (f(g(x))' = f'(y) \cdot g'(x)## where ##y= g(x)## and ##f'## the differentiation according ##y##.
    So ##(- \cos \frac{3}{2} x^2)' = (- \cos y)' \cdot (\frac{3}{2} x^2)' = \sin y \cdot (\frac{3}{2} \cdot 2 \cdot x^{2-1}) = 3x \sin(\frac{3}{2}x^2) ##.
    The argument in the trigonometric function stays as it is.
     
  17. May 17, 2016 #16
    Um, could you explain how I'd integrate that first question a bit more simply, that explanation went right over my head.
     
  18. May 17, 2016 #17

    SteamKing

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    Earlier, you had posted:
    Now, if you wanted to integrate dy = 6 cos (3x) dx to recover y, then obviously you don't integrate the argument of the cosine (which is 3x) by itself.

    The rule for integrating cosine is

    $$\int cos (u) du = sin (u) + C$$

    Here u = 3x, which means that du = 3 dx. If we plug these quantities back into the integral formula, then we get:

    $$\int cos(3x) ⋅ 3\,dx$$

    but we are looking to integrate ##\int 6 ⋅ cos (3x)\,dx##. We can rewrite the integral above as:

    $$2\int cos(3x) ⋅ 3\,dx = 2 sin (3x) + C$$
    which is what we were looking to recover, setting C = 0 of course.
     
  19. May 17, 2016 #18

    SammyS

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    At this point, you should have started a new thread.
     
  20. May 17, 2016 #19

    fresh_42

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    Perhaps I should have written it with ##d##'s.
    $$\frac{d}{dx} f(g(x)) = \frac{d}{dy} f(y) \cdot \frac{d}{dx} g(x)$$
    When you differentiate an antiderivative ##F = \int fdx + const## of a function ##f## you should get ##f## again. So you can always do the test: ##\frac{d}{dx}F(x) = f(x)##.

    If you differentiate ##\sin(3x)## then you get (according to the chain rule) ##\cos(3x) \cdot 3##. The first factor is the derivative of ##\sin## and the second of ##3x##.

    Now you want to integrate ##f(x) = \sin(3x)##, I assume.
    Imagine you have integrated the sine function ##\sin (y)##, so you get as you said ##- \cos (y)##.
    Here ##y = 3x##. So ##\int sin(3x) \, dx ≈ (- \cos (3x))##. But the differentiation for control gives ##\frac{d}{dx} (- \cos(3x)) = 3 \cdot \sin(3x)## according to the chain rule. So there is a correction of a factor ##3## to be made on one side of the ##≈## to make it equal.
     
  21. May 17, 2016 #20

    Mark44

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    @Parsifal1, you should not move on to integration until you o get a better understanding of how the chain rule works in differentiation. Without this understanding, you won't be able to correctly integrate expressions such as sin(3x).

    When you integrate, you should always check your answer by differentiating it. If you get back to the original integrand, your work is probably OK.
     
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