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Finding second derivative of sin y+cos y=x in terms of x

  1. Mar 22, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the second derivative of sin y+cos y=x, giving your answer in terms of x.


    2. Relevant equations
    Implicit derivatives


    3. The attempt at a solution
    ##\sin y+\cos y=x##
    ##\cos y \frac{dy}{dx} -\sin y \frac{dy}{dx}=1##
    ##\frac{dy}{dx}=\frac{1}{\cos y-\sin y}##

    Now, do I use the chain rule by rewriting them as ##(\cos y-\sin y)^{-1}##, or use the quotient rule instead?
     
  2. jcsd
  3. Mar 22, 2014 #2

    Mark44

    Staff: Mentor

    Either one. The chain rule might be a little easier.
     
  4. Mar 22, 2014 #3
    Now that gives:
    ##\frac{d^{2}y}{dx^{2}}=-(\cos y-\sin y)^{-2} (-\sin y \frac{dy}{dx} -\cos y \frac{dy}{dx})##
     
  5. Mar 22, 2014 #4

    Curious3141

    User Avatar
    Homework Helper

    That should simplify to ##\frac{d^2y}{dx^2} = x(\frac{dy}{dx})^3## if you do the proper subs. Actually, it would've been much easier to leave everything on the LHS, and just use product rule and chain rule to get to that.

    But it still doesn't help to put everything in terms of x. Honestly, I cannot see how you can progress from that stage.

    I think you're much better off trying to find y as a function of x, then differentiating twice. This gives a quick answer.

    To do that, think of how you would put ##a\sin x + b\cos x## into the form ##c\sin(x + \theta)##.

    EDIT: There is a way to continue from ##\frac{d^2y}{dx^2} = x(\frac{dy}{dx})^3##. But you have to square the implicit equations for x and the first derivative and add them together to eliminate the y term (apply a trig identity as well). Seems like a bit more effort than just recasting y in terms of x from the start.
     
    Last edited: Mar 23, 2014
  6. Mar 23, 2014 #5
    Well, the form ##c\sin(x+\theta)## worked nicely. :P
     
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