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Homework Help: Derivatives of Trigonometric Equations

  1. Oct 15, 2006 #1
    Alright, moving on to the topic of taking the derivative of a trigonometric equations, I have been given the problem to find the equation of the tangent line to the curve at the given point for:

    y = 1/sinx+cosx at (0,1)

    Now we know that the equation is y-1=m(x-0), so I've tried solving for the derivative and got as far as this:
    y' = (sinx +cosx)d/dx(1) - 1 d/dx (sinx+cosx)
    -------------------------------------
    (sinx+cosx)^2

    = (sinx + cosx)X1 - (1 X (cosx + -sinx) )
    -------------------------------
    (sinx + cosx)^2


    = 2sinx / (sinx + cosx)^2 = 2 cosx / (sinx+cosx)^2

    And then subbing in x=0 and getting 1/1 = 1 and then the equation would be y=x+1, however, I'm not sure if I did this right, or if there is some other way to simplify it that I may have misinterpreted.

    Also, while we are at it, I've just done another question that asks to "find the tangent line to the curve of:"

    y = squareroot(X^2+5) at (2,3)

    I went ahead and solved the derivative and found it to be x/(squareroot(X^2+5)) and subbing in the x=2 got 2/7 as my answer and then just put it into the equation to get y = 2/7X + 17/7, is this correct?

    Thank you anyone, I apologize for asking two questions in one topic haha.
     
  2. jcsd
  3. Oct 15, 2006 #2
    [tex] y = \frac{1}{\sin x + \cos x} [/tex].

    [tex] y' = -\frac{1(\cos x - \sin x)}{(\sin x + \cos x)^{2}} [/tex]. It should be [tex] y = -x +1 [/tex]


    [tex] y' = \frac{x}{\sqrt{x^{2}+5}} [/tex]. You should get [tex] m = \frac{2}{3} [/tex] not [tex] \frac{2}{7} [/tex].
     
    Last edited: Oct 15, 2006
  4. Oct 15, 2006 #3

    radou

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    First, yout function was probably y(x) = 1 / (sinx + cosx). You should watch out for the brackets. Second, what is the derivatice of a constant (in your case 1)?

    Regarding the second function, your answer seems to be ok.
     
  5. Oct 15, 2006 #4

    Office_Shredder

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    For the second function, the square root of 9 is 3, not 7
     
  6. Oct 15, 2006 #5

    radou

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    Right, the equation of the tangent line should be y = 2/3 x + 5/3.
     
  7. Oct 15, 2006 #6
    Yeah I realized that mistake after courtrigrad posted it. I ended up not squarerooting the 5 in X^2+5. So the answer would be 2/3 and it ends up being y= 2/3X + 5/3. I get that question, however, can you explain to me in the 1st problem, why it is -1(cosx-sinx)/(sinx+cosx)^2 because I'm currently going back and reworking that problem out and I'm not sure how you get the -1 part of it, unless I did a step in there that isn't jiving with everything else.
     
  8. Oct 15, 2006 #7

    radou

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    Your function is [tex]f(x) = \frac{1}{\sin x + \cos x} = \frac{g(x)}{h(x)}[/tex]. So, [tex]f'(x) = \frac{g'(x) h(x) - g(x)h'(x)}{h(x)^2}[/tex]. Plug in and solve.
     
  9. Oct 15, 2006 #8
    [tex] y = \frac{1}{\sin x + \cos x} [/tex]

    [tex] f(x) = 1 [/tex]
    [tex] g(x) = \sin x + \cos x [/tex].

    [tex] f'(x) = 0 [/tex]
    [tex] g'(x) = \cos x - \sin x [/tex].

    [tex] \frac{dy}{dx} = \frac{g(x)f'(x)-f(x)g'(x)}{g(x)^{2}} [/tex]
    [tex] \frac{dy}{dx} = \frac{(\sin x + \cos x)(0)-1(\cos x-\sin x)}{(\sin x + \cos x)^{2}} [/tex]
     
  10. Oct 15, 2006 #9
    Oh my goodness, alright I'm following you now, the problem lied within the fact that f'(x) was equal to 0, so I still had the (sinx + cosx) still in the equation. Alright, thanks for your time all of you. I think I'm ready to go back and do some more practice problems to help me grasp this concept fully. :D
     
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