1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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)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


    User Avatar
    Homework Helper

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Derivatives of Trigonometric Equations
  1. Trigonometric equation (Replies: 23)

  2. Trigonometric equation (Replies: 6)