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Derivitives of Trig.

  1. Jul 13, 2007 #1
    1. The problem statement, all variables and given/known data
    1. Sin^2(t) The answer states sin2t
    2. (e^2x)(sin x-cos x) The answer states (e^2x)(3 sin x- cos x)
    3. sec^2(x)-tan^2(x)+cos(x) The answer states -sin x

    2. Relevant equations
    1.product rule
    2.product rule
    3. sum and diff. rule?

    3. The attempt at a solution
    1. (sin t)(sin t)'+(sin t)(sin t)'=(sin t)(cos t)+(sin t)(cos t)=2cos (t) sin (t)
    2. (e^2x)(cos x+sin x)+ (e^2x)(sin x-cos x)=(e^2x)(sin 2x)
    3. (sec^2(x)*tan^2(x))-sec^4(x)-sin x. I just don't know how to get rid of the first and second functions.
  2. jcsd
  3. Jul 13, 2007 #2


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    This is correct; you need to use a trig identity to get it in the form of the solution. Note that, alternatively, you could have used the chain rule for this question.
    This isn't quite correct; what is the derivative of e^(2x)?
    I'm not sure what you've done here. You want to differentiate the original term by term, so you need to find the derivatives of sec^2x and tan^2x. Can you write these as quotients and use the quotient rule?
  4. Jul 13, 2007 #3
    For 1. I tried the chain rule(sin x inside, square outside), and it got me 2cos(t) which is worse than my first answer.
    For 2. I assumed e never changes. And I never came across a problem where e changes and I am not too sure on how to do that. Is it 2 e^1?
    For 3. If I am understanding this correctly, Instead of the original, say it as: sec^2(x)+cos(x)/tan^2(x) then apply the quotient rule?
    If I am not correct, please guide me to the answer(s).
  5. Jul 13, 2007 #4


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    You made a mistake using the chain rule. Recall that [tex]\frac{d}{dx}((\sin x)^2)=2\sin x \cdot\frac{d}{dx}(\sin x)=2\sin x\cos x [/tex]. Now, can you spot the trig identity that will reduce the answer to the required form?
    [tex]\frac{d}{dx}(e^{ax})=ae^{ax}[/tex] for some constant a.
    No, I meant write the expression as [tex]\frac{1}{\cos^2x}-\frac{\sin^2x}{\cos^2x}+\cos x [/tex] and differentiate each term, using the quotient rule to differentiate the first two terms.
    Last edited: Jul 13, 2007
  6. Jul 14, 2007 #5
    For 1, I am really not seeing how it becomes sin2t from 2sinx cosx.
    For 2, I got it right, it was the derivative of e^2x that messed me up. thanks cristo.
    For 3, I know ultimately the third term will be left and the other two will cancel each other out. This is what i got(by following the advice):
    f'(x)=(cosx^2-sinx^2/(cosx^2)^2)-(cosx^4+sinx^4)/(cosx^2)^2)-sinx. I know the exponents does not cancel in this step.

    For 1 and 3, I know there is a step that I do not have any knowledge of(or i did a step wrong), thanx in advance.
  7. Jul 14, 2007 #6


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    Ok, it's just a trig identity you should remember. The identity is [itex]\sin(2x)=2\sin x\cos x[/itex]. It is derived from the multiple angle formula [itex]\sin(a+b)=\sin a\cos b+\cos a\sin b[/itex] by setting a=b
    You're welcome
    Ok, let's look at each term individually, since you have used the quotient rule incorrectly on (at least) the first term.

    We want [tex]\frac{d}{dx}\left(\frac{1}{\cos^2x}\right)[/tex]. Using the quotient rule, this is equal to [tex]\frac{\cos^2x\cdot\frac{d}{dx}(1)-1\cdot\frac{d}{dx}(\cos^2x)}{(\cos^2x)^2}=
    \frac{2\cos x\sin x}{\cos^4x}=\frac{2\sin x}{\cos^3x}[/tex]

    Now, try to do the second term (sin^2x/cos^2x) again. I'm not sure how you got what you did first time.
  8. Jul 14, 2007 #7
    3. sinx^2/cosx^2

    (cosx^2)(2 cos x sin x)+(2 cos x sin x)(sinx^2)

    (2 cosx^2*cosx^3*sin x cosx^2)+(2sinx^2*cosx sinx^2*sinx^2) distribution
    Is this right so far?
  9. Jul 14, 2007 #8
    For #3,

    sec^2(x) - tan^2(x) = 1.

    the rest is easy.
  10. Jul 14, 2007 #9


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    That is indeed a quicker way of doing the question (I can't believe I didn't notice that!) cd246, this comes about from the identity sin^2x+cos^2x=1, and greatly simplifies the work needed.
  11. Jul 14, 2007 #10
    I think I get what u 2 are saying... since sec^2-tan^2 equals 1, it would be 1+cosx, take the derivative and I get -sinx?
    Last edited: Jul 14, 2007
  12. Jul 14, 2007 #11


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    Yup, that's correct.
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