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Thread: Properities of a derivative

  1. #1
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    Properities of a derivative

    If $\displaystyle g(t)= sin(f(-t, t, 2t))$ find $\displaystyle g'(t)$. im not sure exactly how to approach this one.
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    Quote Originally Posted by Scopur View Post
    If $\displaystyle g(t)= sin(f(-t, t, 2t))$ find $\displaystyle g'(t)$. im not sure exactly how to approach this one.
    use the chain rule

    in single variable it says: $\displaystyle \frac d{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$

    can you generalize that to this case?
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    Im more so having problems with the sin in front... if it wasn't there id know what to do.

    If it were $\displaystyle g(t)=(f(-t, t, 2t))$ find $\displaystyle g'(t)$ i would just set $\displaystyle g=f(c(t))$ and solve
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Scopur View Post
    Im more so having problems with the sin in front... if it wasn't there id know what to do.

    If it were $\displaystyle g(t)=(f(-t, t, 2t))$ find $\displaystyle g'(t)$ i would just set $\displaystyle g=f(c(t))$ and solve
    look at it this way: how would you find the derivative of something like $\displaystyle y = \sin (f(x))$? if that's too general, imagine $\displaystyle y = \sin (\cos x)$. how would you deal with differentiating that?
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  5. #5
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    Hello,

    Let u, v, w be functions of t.

    $\displaystyle \frac{df(u,v,w)}{dt}=\frac{\partial f(u,v,w)}{\partial u} \cdot \frac{du}{dt}+\frac{\partial f(u,v,w)}{\partial v} \cdot \frac{dv}{dt}+\frac{\partial f(u,v,w)}{\partial w} \cdot \frac{dw}{dt}$
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    I still dont really get it.. heres my stab it

    $\displaystyle f(c(t)) = sin(f(-t,t,2t))$
    $\displaystyle g'(t)= cos(f(-t, t, 2t) \cdot c('t)$


    Is this right.... or am i completely wrong.. i really need to see a computation of this the above examples really haven't helped me
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Scopur View Post
    I still dont really get it.. heres my stab it

    $\displaystyle f(c(t)) = sin(f(-t,t,2t))$
    $\displaystyle g'(t)= cos(f(-t, t, 2t) \cdot c('t)$


    Is this right.... or am i completely wrong.. i really need to see a computation of this the above examples really haven't helped me
    $\displaystyle g(t) = \sin [f(-t,t,2t)]$

    $\displaystyle \Rightarrow g'(t) = \cos [f(-t,t,2t)] \cdot \frac {df(-t,t,2t)}{dt}$, by the chain rule

    now follow what Moo said to find $\displaystyle \frac {df(-t,t,2t)}{dt}$
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    okay okay i get it thank you.
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