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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- Oct 18th 2008, 11:15 AMScopurProperities 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.

- Oct 18th 2008, 12:17 PMJhevon
- Oct 18th 2008, 12:25 PMScopur
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 - Oct 18th 2008, 12:55 PMJhevon
- Oct 18th 2008, 01:11 PMMoo
Hello,

Let u, v, w be functions of t.

$\displaystyle \frac{df(u,v,w)}{dt}=\frac{\partial f(u,v,w)}{\partial u} \cdot \frac{du}{dt}+\frac{\partial f(u,v,w)}{\partial v} \cdot \frac{dv}{dt}+\frac{\partial f(u,v,w)}{\partial w} \cdot \frac{dw}{dt}$ - Oct 18th 2008, 01:31 PMScopur
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 - Oct 18th 2008, 01:42 PMJhevon
- Oct 18th 2008, 01:44 PMScopur
okay okay i get it thank you.