# Properities of a derivative

• Oct 18th 2008, 11:15 AM
Scopur
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.
• Oct 18th 2008, 12:17 PM
Jhevon
Quote:

Originally Posted by Scopur
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?
• Oct 18th 2008, 12:25 PM
Scopur
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 PM
Jhevon
Quote:

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 $\displaystyle g(t)=(f(-t, t, 2t))$ 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?
• Oct 18th 2008, 01:11 PM
Moo
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 PM
Scopur
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 PM
Jhevon
Quote:

Originally Posted by Scopur
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 $\displaystyle \frac {df(-t,t,2t)}{dt}$
• Oct 18th 2008, 01:44 PM
Scopur
okay okay i get it thank you.