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malevolence19 Group TitleBest ResponseYou've already chosen the best response.0
You need the product rule right?
 one year ago

malevolence19 Group TitleBest ResponseYou've already chosen the best response.0
And the chain rule, do you know both of these?
 one year ago

v.s Group TitleBest ResponseYou've already chosen the best response.0
i did like 5 questions like this got all of them wrong
 one year ago

malevolence19 Group TitleBest ResponseYou've already chosen the best response.0
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

malevolence19 Group TitleBest ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=derivative+of+e+*+a+*+x+*+sin(Bx) It's not...
 one year ago
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