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Math Help - Derivative help

  1. #1
    Member OnMyWayToBeAMathProffesor's Avatar
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    Derivative help

    Hello, so I have this problem that I am not completely sure on, here it is:

    Let f be the real-valued function defined by f(x)=sin^3x + sin^3|x|

    (a) Find f'(x) for x>0.
    (b) Find f'(x) for x<0.
    (c) Determine whether f(x) is continuous at x=0. Justify your answer
    (d) Determine whether the derivative of f(x) exists at x=0. Justify your answer.

    So I know how to do power rule and I have memorized some standards derivative values. But I do not know about raising sin to the third power. If x was cubed I would be able to do it, but not sin.

    For b it would be different because the absolute value of x = \sqrt(x^2). I do not know how to take the derivative of that.

    For c I said no because sin is never continuous at 0?

    Finally for d i out that it does not because f(x) is not continuous at 0.

    ANY help would be greatly appreciated once again. Thanks.
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  2. #2
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    So I know how to do power rule and I have memorized some standards derivative values. But I do not know about raising sin to the third power. If x was cubed I would be able to do it, but not sin.
    Are you familiar with the chain rule? You will need to use that here.

    Let f be the real-valued function defined by f(x)=\sin^3x+\sin^3\lvert x\rvert

    (a) Find f'(x)\text{ for }x>0.
    For x>0,\;\lvert x\rvert=x. So,

    f'(x)=\frac d{dx}\left[\sin^3x+\sin^3x\right]

    =\frac d{dx}\left[2\sin^3x\right]

    =\left(6\sin^2x\right)\frac d{dx}\left[\sin x\right]

    =6\sin^2x\cos x.

    (b) Find f'(x)\text{ for }x<0.
    When x<0,\;\lvert x\rvert=-x. So, remembering that sine is an odd function ( \sin(-a)=-\sin a), we have

    f'(x)=\frac d{dx}\left[\sin^3x+\sin^3(-x)\right]

    =\frac d{dx}\left[\sin^3x+(-\sin x)^3\right]

    =\frac d{dx}\left[\sin^3x-\sin^3x\right]

    =0.

    (c) Determine whether f(x) is continuous at x=0. Justify your answer
    ...
    For c I said no because sin is never continuous at 0?
    Why would you think that? Sine is continuous over the entire real line.

    \lvert x\rvert is everywhere continuous, \sin x is everywhere continuous, and x^3 is everywhere continuous. The composition of two continuous functions is itself continuous, so \sin^3x and \sin^3\lvert x\rvert are both continuous. Finally, the sum of two continuous functions is continuous, so f(x) is everywhere continuous (including x=0).

    If you like, you could also use the definition of continuity at a point by showing that \lim_{x\to0}f(x)=f(0) (to do this, consider the one-sided limits).

    (d) Determine whether the derivative of f(x) exists at x=0. Justify your answer.
    ...
    Finally for d i out that it does not because f(x) is not continuous at 0.
    See above. f is continuous at x=0, and in fact, f is also differentiable at x=0. Consider the left- and right-sided limits.

    \lim_{x\to0^-}\frac{f(x)-f(0)}x

    =\lim_{x\to0^-}0=0

    and

    \lim_{x\to0^+}\frac{f(x)-f(0)}x

    \lim_{x\to0^+}6\sin^2x\cos x=6\sin^20\cos0=0.

    Both one-sided limits are equal, so the limit f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}x exists.
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  3. #3
    Member OnMyWayToBeAMathProffesor's Avatar
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    I am sorry i was thinking of \frac{sinx}{x}, which is not continuous at 0. and how does |x|=-x for x<0 ? I thought if x=-6 which would be |-6| which is 6, correct?
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  4. #4
    Member OnMyWayToBeAMathProffesor's Avatar
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    nvm, i see what happened. thank you very much for a clear and concise explanation of the problem and the solution.
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