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    I have to use the chain rule to differentiate y=sin(t^2+9t+11)

    Is it just U=(t^2+9t+11) and Y=sin(U) ??
    Is this right? Why do i even have to use the chain rule for this case?
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    (Original post by ErniePicks)
    I have to use the chain rule to differentiate y=sin(t^2+9t+11)

    Is it just U=(t^2+9t+11) and Y=sin(U) ??
    Is this right? Why do i even have to use the chain rule for this case?
    Yes, that's correct, then we have that \displaystyle \frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}u} \times \frac{\mathrm{d}u}{\mathrm{d}t} = \cos (u) \times \frac{\mathrm{d}}{\mathrm{d}t} (t^2 + 9t +11)
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    Yeah that's correct, why would you not use the chain rule? It is a function in a function.
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    (Original post by Dinasaurus)
    Yeah that's correct, why would you not use the chain rule? It is a function in a function.
    I got 2t+9(cos(t^2+9t+11)) as the answer though, it doesn't look right...
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    (Original post by ErniePicks)
    I got 2t+9(cos(t^2+9t+11)) as the answer though, it doesn't look right...
    Most chain rule answers look messier than the original function given.
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    (Original post by ErniePicks)
    I got 2t+9(cos(t^2+9t+11)) as the answer though, it doesn't look right...
    If you mean (2t+9)\cos (t^2 + 9t+11) \neq 2t + 9(\cos (t^2 + 9+ 11)) (if you mean the former) then you're correct, if you mean the latter, then you're wrong.
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    you are good human beings
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    (Original post by ErniePicks)
    I got (2t+9)(cos(t^2+9t+11)) as the answer though, it doesn't look right...
    just remember to bracket that term but other than that its perfect


    doesnt look "right" because sin(t^2+9t+11) is a weird looking function :laugh:

    Edit: i think there is an echo whoops
 
 
 
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