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Derivatives of trigonometric functions

  1. Oct 4, 2008 #1
    Hi, i'm working with finding the derivatives of trigonometric functions but i'm not confidant with some of my answers. if someone would go over these derivatives i would appreciate it. thanks in advance!

    determine [tex]\frac{dy}{dx}[/tex] . do not simplify.

    question 1
    y = sec [tex]\sqrt[3]{x}[/tex]

    my answer:
    y1 = sec [tex]\sqrt[3]{x}[/tex] tan[tex]\sqrt[3]{x}[/tex]
    = sec [tex]\sqrt[3]{x}[/tex] tan[tex]\sqrt[3]{x}[/tex] [tex]\frac{1}{3}[/tex] x[tex]^{- \frac{2}{3}}[/tex]

    = [tex]\frac{sec \sqrt[3]{x} tan \sqrt[3]{x}}{ 3\sqrt[3]{x^{2}}}[/tex]

    question 2

    y = 4cos[tex]^{3}[/tex] ([tex]\pi[/tex] x)

    my answer:

    y1 = 12(-sin [tex]^{2}[/tex] [tex]\pi[/tex] x) [tex]\pi[/tex]

    = -12([tex]\pi[/tex] sin [tex]^{2}[/tex] [tex]\pi[/tex] x

    question 3

    y = 2x([tex]\sqrt{x}[/tex] - cot x)

    my answer:
    y1 = 2([tex]\sqrt{x}[/tex] - cot x) + (2x) [tex]\frac{1}{2}[/tex] x[tex]^{- \frac{1}{2}}[/tex] - (-csc [tex]^{2}[/tex] x))

    = 2([tex]\sqrt{x}[/tex] - cot x) + (2x) [tex]\frac{csc^{2}x}{2\sqrt{x}}[/tex]
    = 2([tex]\sqrt{x}[/tex] - cot x) + [tex]\frac{x csc^{2}x}{\sqrt{x}}[/tex]

    question 4

    y = tan [tex]^{2}[/tex] (cos x)

    my answer
    y1 = 2sec[tex]^{2}[/tex] (-sinx)

    question 5
    y = [tex]\frac{1}{1 + tanx}[/tex]

    my answer
    y1 = [tex]\frac{1}{sec^{2}x}[/tex]

    question 6
    sinx + siny = 1
    cos x + cos y [tex]\frac{dy}{dx}[/tex] = 0

    [tex]\frac{dy}{dx}[/tex] = - [tex]\frac{cosx}{cosy}[/tex]


    I'm not sure how i did with these. if someone could overlook them i'd be very greatful.
     
  2. jcsd
  3. Oct 4, 2008 #2

    Dick

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    You started going wrong in 2. Take a hard look at some examples of using the chain rule. Here for example it would say (f(x)^3)'=3*f(x)^2*f'(x). Apply this to the case where f(x)=cos(pi*x).
     
  4. Oct 4, 2008 #3
    so would question 2 be...

    4 cos[tex]^{2}[/tex] ([tex]\pi[/tex] x)

    12 cos [tex]^{2}[/tex] ([tex]\pi[/tex] x) (-sin ([tex]\pi[/tex] x)) ([tex]\pi[/tex])

    (-sin ([tex]\pi[/tex] x)) 12 [tex]\pi[/tex] cos [tex]^{2}[/tex] ([tex]\pi[/tex] x)
     
  5. Oct 4, 2008 #4

    Dick

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    That's it.
     
  6. Oct 4, 2008 #5
    great thank you!!!

    as for question 3
    y = 2x( [tex]\sqrt{x}[/tex] - cotx)

    my second attempt at an answer:
    y1 = 2( [tex]\sqrt{x}[/tex] - cotx) + 2x([tex]\frac{1}{2}[/tex] x[tex]^{- \frac{1}{2}}[/tex] + csc[tex]^{2}[/tex] x
    = 2( [tex]\sqrt{x}[/tex] - cotx) + [tex]\sqrt{x}[/tex] + 2x csc[tex]^{2}[/tex] x


    is this anywhere near correct?
    thanks

    i am not sure how to get on with question 4: y = tan [tex]^{2}[/tex] (cosx)

    thank you for the help!
     
    Last edited: Oct 4, 2008
  7. Oct 4, 2008 #6

    Dick

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    That's also correct. You could combine the square roots by expanding out, but that doesn't make it incorrect. For 4, it might help to rewrite it a little. Let's define sqr(x)=x^2, ok? Then that is sqr(tan(cos(x))). If you use the chain rule twice you can show that (f(g(h(x)))'=f'(g(h(x))*g'(h(x))*h'(x). Do you see how that's working? You keep taking the derivative of the outside function evaluated at the inside function times the derivative of the inside function.
     
  8. Oct 4, 2008 #7
    alright...i'm not sure i got this one but here it goes:

    y = tan [tex]^{2}[/tex] (cos x)

    y1 = 2tan cosx + 2 sec[tex]^{2}[/tex]cosx - sin tan[tex]^{2}[/tex]

    how does it look??? :|
     
  9. Oct 4, 2008 #8
    also are question five and six okay? i dont see how i can come up with any other solutions.
     
  10. Oct 4, 2008 #9

    Dick

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    It looks kind of incoherent. You can either use full tex stuff or you can do what I usually do and try to approximate it with lots of characters and parenthesis. But I don't understand that at all.
     
  11. Oct 4, 2008 #10

    Dick

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    I think 6 is ok. 5 is awful. (1/f(x))' is not equal to 1/(f'(x)), is it?
     
  12. Oct 5, 2008 #11
    okey i believe i figured question 5 out:

    y = [tex]\frac{1}{1+tanx}[/tex]

    y1 = (1 + tanx)[tex]^{-1}[/tex]
    = (-1)(1 + tanx)(sec[tex]^{-2}[/tex] sec [tex]^{2}[/tex] x
    = - [tex]\frac{(sec^2)x}{(1 + tanx)^2}[/tex]
    I am still unsure where i am going with question 4 but here is my zillionth attempt :D

    y = tan[tex]^{2}[/tex](cos x)

    my answer:

    y1 = 2tan(cosx)sec[tex]^{2}[/tex]x(cosx)(-sinx)
    y1 = -2sin x tan(cosx) sec[tex]^{2}[/tex](cosx)


    thank you. i appreciate all the help! :D
     
    Last edited: Oct 5, 2008
  13. Oct 5, 2008 #12

    Dick

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    They both look correct. You are improving a lot!
     
  14. Oct 5, 2008 #13
    Thank you sooo much! couldn't have done that without you! THANKS for alllll the help! :D
     
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