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Math Help - find y' if y=(sinx)^2 + 2^(sinx)

  1. #1
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    find y' if y=(sinx)^2 + 2^(sinx)

    find y' if y=(sinx)^2 + 2^{sinx}

    I don't really know which definitions/properties to use in this instance, but what I've tried is:

    Dx a^{x} = a^{x}\ln{a}

    First, I separated the two terms and took the derivative of both

    Dx (sinx)^2 = (sinx)^2\ln{sinx}

    Dx 2^{sinx} = 2(sinx)(cosx) (Chain Rule)

    Giving: (sinx)^2\ln{sinx} +  2(sinx)(cosx)

    Any chance that I did this right? If not, am I using the wrong properties for the first term? Thanks!!
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  2. #2
    Newbie Mr. Engineer's Avatar
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    Quote Originally Posted by littlejodo View Post
    find y' if y=(sinx)^2 + 2^{sinx}

    I don't really know which definitions/properties to use in this instance, but what I've tried is:

    Dx a^{x} = a^{x}\ln{a}

    First, I separated the two terms and took the derivative of both

    Dx (sinx)^2 = (sinx)^2\ln{sinx}

    Dx 2^{sinx} = 2(sinx)(cosx) (Chain Rule)

    Giving: (sinx)^2\ln{sinx} +  2(sinx)(cosx)

    Any chance that I did this right? If not, am I using the wrong properties for the first term? Thanks!!
    Okay, for the first term, (sinx)^2, you would simply use the chain rule and you would get
    2(sinx)(cosx)
    The take the derivative of the second term using
    a^{x} = a^{x}\ln{a}
    and you would get as a final answer
    cosx(2(sinx)+2^{sinx}ln2)
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  3. #3
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    It looks like you must have had a cosx in both terms in order to be able to factor it out. Is this because in the second term after getting 2^(sinx)ln2 you have to get the derivative of the exponent, sinx, which is cosx?

    Thanks!!!!
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by littlejodo View Post
    It looks like you must have had a cosx in both terms in order to be able to factor it out. Is this because in the second term after getting 2^(sinx)ln2 you have to get the derivative of the exponent, sinx, which is cosx?

    Thanks!!!!
    yes, this is due to the chain rule

    for a > 0

    \frac d{dx}a^x = a^x \ln a

    so, by the chain rule, \frac d{dx}a^u = u'a^u \ln a

    where u is a function of x


    This follows from \frac d{dx} f(g(x)) = f'(g(x)) \cdot g'(x) (the Chain Rule)

    in the second rule i gave you, f(x) = a^x and g(x) = u
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