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Math Help - mastery test; derivative question

  1. #1
    Member mybrohshi5's Avatar
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    mastery test; derivative question

    im studying for this test i have to take on monday called the mastery test. its all derivatives and im studying this practice sheet my teacher posted online but i think i found an error on one. can someone check it please.

    y=x^sin(x)

    find y'

    y'=(ln(x))(x^sin(x))(cos(x))

    this is what i got ^^^^

    but on the answer sheet of the practice sheet it says that the answer is:

    [cosx(lnx) + sinx(1/x)] (x^sinx)

    which one is correct?

    thank you
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  2. #2
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    Quote Originally Posted by mybrohshi5 View Post
    im studying for this test i have to take on monday called the mastery test. its all derivatives and im studying this practice sheet my teacher posted online but i think i found an error on one. can someone check it please.

    y=x^sin(x)

    find y'

    y'=(ln(x))(x^sin(x))(cos(x))

    this is what i got ^^^^

    but on the answer sheet of the practice sheet it says that the answer is:

    [cosx(lnx) + sinx(1/x)] (x^sinx)

    which one is correct?

    thank you
    The answer sheet is correct.

     y = x^{\sin(x)}

     \ln(y) = \ln(x^{\sin(x)})

     \therefore \ln(y) = \sin(x) \ln (x)

    Differentiate this implicitly using the chain rule on the LHS, and using the product rule on the RHS:

    The Chain Rule:

     \frac{d}{dx} f(y) = \frac{d f(y)}{dy} \times \frac{dy}{dx}

    The Product Rule:

     \frac{d}{dx} f(x) \times g(x) = f (x) \times \frac{d g(x)}{dx} + g(x) \times \frac{d f(x)}{dx}
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  3. #3
    Member mybrohshi5's Avatar
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    oopps. thank you. i just saw there was an x in the power and was using the theorem or whatever it is you can use like for y=5^x ..... y'=ln5(5^x)(1) but its different when there is an x raising an x to a power haha. silly me.

    thanks again
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  4. #4
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    Quote Originally Posted by mybrohshi5 View Post
    oopps. thank you. i just saw there was an x in the power and was using the theorem or whatever it is you can use like for y=5^x ..... y'=ln5(5^x)(1) but its different when there is an x raising an x to a power haha. silly me.

    thanks again
    It's not exactly a theorem, but when you have y = 5^x the derivative is obtained using the same procedure Mush used.

    It's best not to remember some formula when tackling these types of derivatives but to remember the procedure of taking the natural log of both sides, differentiating both sides (using the product rule or chain rule if necessary) and then rearranging to find \frac{dy}{dx}
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