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steel11

  • 3 years ago

derivative of 2xsin(1/x) my answer i put, which is wrong 2sinx^-1*cosx^-1*-1x^-2 can anybody give me correct answer or find where i went wrong? im doing the chain rule, and i think you have to do twice, which i did. btw, i brought the x to the numerator with -1 as exponent.

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  1. RadEn
    • 3 years ago
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    hi uzu :)

  2. uzumakhi
    • 3 years ago
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    do u know abut product derivation d\dx(a*b) = a d\dx(b) + b d\dx(a)

  3. steel11
    • 3 years ago
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    product rule? yes

  4. uzumakhi
    • 3 years ago
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    try to use it

  5. steel11
    • 3 years ago
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    I would like to do this with chain rule. If thats possible, which I think it is.

  6. uzumakhi
    • 3 years ago
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    try to use my method and tell me the answer u got

  7. across
    • 3 years ago
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    Let\[f(x)=2x\sin\left(\frac1x\right).\]Then\[f'(x)=2\sin\left(\frac1x\right)+2x\cos\left(\frac1x\right)\cdot\left(-\frac1{x^2}\right)=2\left[\sin\left(\frac1x\right)-\frac1x\cos\left(\frac1x\right)\right].\]

  8. steel11
    • 3 years ago
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    thank you across. you used product rule and chain rule to solve right?

  9. steel11
    • 3 years ago
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    guess uzumaki was right on having to use product rule. ._.

  10. uzumakhi
    • 3 years ago
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    that what i am talking about thanx @across

  11. steel11
    • 3 years ago
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    =]

  12. across
    • 3 years ago
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    Yes. To be more specific, let me break it down for you: Essentially, you have this:\[f(x)=g(x)h(j(x)),\]where\[g(x)=2x\\h(x)=\sin x\\j(x)=\frac1x.\]Then\[f'(x)=g'(x)h(j(x))+g(x)h'(j(x))j'(x),\]and since\[g'(x)=2\\h'(x)=\cos x\\j'(x)=-\frac1{x^2},\]we have that\[f'(x)=2\cos\left(\frac1x\right)+2x\cos\left(\frac1x\right)\cdot\left(\frac1{x^2}\right).\]

  13. across
    • 3 years ago
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    I forgot the negative sign in the last term, but you get the idea. I used the product and chain rules.

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