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nickymarden

  • 2 years ago

derive the funtion.

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  1. nickymarden
    • 2 years ago
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  2. satellite73
    • 2 years ago
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    \[f(x)=\frac{\sin^2(x^2+5)}{\sqrt{x^2+1}}\]? like that? will be an algebra nightmare your math teacher must hate you

  3. nickymarden
    • 2 years ago
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    haha, that's right. I was supposed t do it in class, but i got to the middle and got fed up. Can you help me? please..

  4. satellite73
    • 2 years ago
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    \[(\frac{f}{g})'=\frac{gf'-f'g}{g^2}\] with \[f(x)=\sin^2(x^2+5),g(x)=\sqrt{x^2+1}\] we need both \(f'\) and \(g'\) and both require the chain rule

  5. satellite73
    • 2 years ago
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    so first we note that \(f\) is actually a composition of three functions, \(x^2+5\) , sine and the squaring function so good \[f'(x)=2\sin(x^2+5)\times \cos(x^2+5)\times 2x=4x\sin(x^2+5)\cos(x^2+5)\]

  6. satellite73
    • 2 years ago
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    and also by the chain rule \[g'(x)=\frac{2x}{2\sqrt{x^2+1}}=\frac{x}{\sqrt{x^2+1}}\] so far so good?

  7. nickymarden
    • 2 years ago
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    yeees. :)

  8. satellite73
    • 2 years ago
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    now we have to put this mess together for \[(\frac{f}{g})'=\frac{gf'-f'g}{g^2}\] the denominator is easy enough, it is just \(x^2+1\) but the numerator is going to be a colossal mess i guess we can just write it down

  9. satellite73
    • 2 years ago
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    numerator is \[\sqrt{x^2+1}\times 4x\sin(x^2+5)\cos(x^2+5)-\sin^2(x^2+5)\times \frac{x}{\sqrt{x^2+1}}\]

  10. satellite73
    • 2 years ago
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    and so your whole mess is \[\frac{\sqrt{x^2+1}\times 4x\sin(x^2+5)\cos(x^2+5)-\sin^2(x^2+5)\times \frac{x}{\sqrt{x^2+1}}}{x^2+1}\]

  11. satellite73
    • 2 years ago
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    now you can get rid of the compound fraction by multiplying top and bottom by \(\sqrt{x^2+1}\) to get \[\frac{(x^2+1)\times 4x\sin(x^2+5)\cos(x^2+5)-\sin^2(x^2+5)\times x}{(x^2+1)\sqrt{x^2+1}}\] and maybe you can even clean this mess up with some more algebra by multiplying out in the numerator or something, but my algebra is not that good

  12. nickymarden
    • 2 years ago
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    omg, if i could give you a million medals i would.

  13. satellite73
    • 2 years ago
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    there is also an annoying trig identity you can use since \[2\sin(x)\cos(x)=\sin(2x)\] so if you want to be really clever you can replace \[4x\sin(x^2+5)\cos(x^2+5)\] by \[2x\sin(2x^2+10)\] but i wouldn't

  14. satellite73
    • 2 years ago
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    one is more than enough yw

  15. satellite73
    • 2 years ago
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    oh also you can write the denominator as \((x^2+1)^{\frac{3}{2}}\) but that strikes me as unnecessary also

  16. nickymarden
    • 2 years ago
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    but he wants it like that lol thank you :)

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