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v.s

Find y' and y''. y = eαx sin βx

  • one year ago
  • one year ago

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  1. malevolence19
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    You need the product rule right?

    • one year ago
  2. malevolence19
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    And the chain rule, do you know both of these?

    • one year ago
  3. v.s
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    i did like 5 questions like this got all of them wrong

    • one year ago
  4. malevolence19
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    Well first off, the product rule. If you have two function multiplied together (in this case: x and sin(Bx)) then you apply: \[\frac{d}{dx}f(x)*g(x)=f'(x)g(x)+f(x)g'(x)\] As for the chain rule, when you apply it to sin(Bx) you get: \[\frac{d}{dx}\sin( \beta x)=\beta \cos(\beta x)\] And you also know that: \[\frac{d}{dx}(af(x))=a \frac{d}{dx}f(x)\] So applying all these you get: \[y'=\frac{d}{dx}(e \alpha x \sin(\beta x))=e \alpha \frac{d}{dx}x \sin (\beta x)=e \alpha \left[ \frac{d}{dx}(x)\sin(\beta x)+x \frac{d}{dx}\sin(\beta x)\right]\] \[=e \alpha \left[ \sin(\beta x) + \beta x \cos(\beta x)\right]\] Now apply those rules again to get y''.

    • one year ago
  5. v.s
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    its wrong

    • one year ago
  6. malevolence19
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    http://www.wolframalpha.com/input/?i=derivative+of+e+*+a+*+x+*+sin(Bx) It's not...

    • one year ago
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