Properities of a derivative

**Scopur** · October 18th 2008, 11:15 AM

If $g(t)= sin(f(-t, t, 2t))$ find $g'(t)$ . im not sure exactly how to approach this one.

**Jhevon** · October 18th 2008, 12:17 PM

Originally Posted by Scopur

If $g(t)= sin(f(-t, t, 2t))$ find $g'(t)$ . im not sure exactly how to approach this one.

use the chain rule

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

can you generalize that to this case?

**Scopur** · October 18th 2008, 12:25 PM

Im more so having problems with the sin in front... if it wasn't there id know what to do.

If it were $g(t)=(f(-t, t, 2t))$ find $g'(t)$ i would just set $g=f(c(t))$ and solve

**Jhevon** · October 18th 2008, 12:55 PM

Originally Posted by Scopur

Im more so having problems with the sin in front... if it wasn't there id know what to do.

If it were $g(t)=(f(-t, t, 2t))$ find $g'(t)$ i would just set $g=f(c(t))$ and solve

look at it this way: how would you find the derivative of something like $y = \sin (f(x))$ ? if that's too general, imagine $y = \sin (\cos x)$ . how would you deal with differentiating that?

**Moo** · October 18th 2008, 01:11 PM

Hello,

Let u, v, w be functions of t.

$\frac{df(u,v,w)}{dt}=\frac{\partial f(u,v,w)}{\partial u} \cdot \frac{du}{dt}+\frac{\partial f(u,v,w)}{\partial v} \cdot \frac{dv}{dt}+\frac{\partial f(u,v,w)}{\partial w} \cdot \frac{dw}{dt}$

**Scopur** · October 18th 2008, 01:31 PM

I still dont really get it.. heres my stab it

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

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

**Jhevon** · October 18th 2008, 01:42 PM

Originally Posted by Scopur

I still dont really get it.. heres my stab it

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

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

$g(t) = \sin [f(-t,t,2t)]$

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

now follow what Moo said to find $\frac {df(-t,t,2t)}{dt}$

**Scopur** · October 18th 2008, 01:44 PM

okay okay i get it thank you.

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