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Homework Help: Trigonometry differentiation problem

  1. Dec 22, 2009 #1
    1. The problem statement, all variables and given/known data
    Differentiate the following
    [tex]cos^2 x[/tex]

    3. The attempt at a solution

    \frac{d(cos^2 x)}{dx}

    = 2 (-sin^{2-1} x).\frac{dx}{dx}

    = 2 (-sin x).1

    = -2 sin x


    But the answer given is-
    -2 cos x sin x

    Can you please tell me what's wrong !!!!!!
  2. jcsd
  3. Dec 22, 2009 #2


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    Homework Helper

    You should be more careful in applying the chain rule. There are many formulations, the way I prefer it is to let u = cos(x). Then you are asked for d(u2(x))/dx. According to the chain rule, this is
    [tex]\frac{d}{dx} u(x)^2 = \frac{d(u^2)}{du} \cdot \frac{du}{dx}[/tex]
    The first factor is 2u = 2 sin(x), but then you need to multiply du/dx, not dx/dx.
  4. Dec 22, 2009 #3


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    Or another rule of thumb which I prefer is to always take the derivative of what's inside... I'll show you what I mean:

    If we are to take the derivative of [itex]x^n[/itex] we get [itex]nx^{n-1}[/itex]
    but behind the scenes you're also taking the derivative of x, but since this is 1, multiplying by 1 doesn't change anything.

    Now if we let x be anything else, such as sinx then we have [itex](sinx)^n[/itex] and taking the derivative of this is the same as before, but this time the derivative of what's inside (in this case, sinx) is cosx so this makes a difference: [itex]n(sinx)^{n-1}.cosx[/itex]

    But you also need to dig deeper and keep taking the derivative of what's inside, if there is something further inside.

    e.g. [itex]\left(sin(1/x)\right)^n[/itex]
    Again, take the derivative by first applying the power rule, and then taking the derivative of sin(1/x) by using the trigo change, so cos(1/x) multiplies on the outside, but finally you also need to take the derivative of 1/x (this was also done behind the scenes in the 2nd example, we took the derivative of x, but again that's 1) so altogether we get:


    And you can of course simplify. Anyway, I know this is terribly set out but I hope you understand it :smile:
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