# Derivative qus.

• Apr 29th 2006, 10:22 AM
Derivative qus.
Hey guys,

Finding hard to solve these derivative:

1) $q = 2e^{-t/2}cos2t$ (don't know how to deal with the index -t/2)

2) $x = e^{2t}t^3(2-t)^4$ (this seem confusing as don't know what rule to apply)

3) $e^{x+y} + sin3x = 0$ (not to sure about index agian)

I hope someone can help. Thanks

• Apr 29th 2006, 10:37 AM
CaptainBlack
Quote:

1) $q = 2e^{-t/2}\cos(2t)$ (don't know how to deal with the index -t/2)

$
\frac{dq}{dt}=2 \left[\left(\frac{d}{dt}e^{-t/2}\right)\cos(2t)+e^{-t/2}\left(\frac{d}{dt} \cos(2t)\right)\right]
$

so using the chain rule to do the derivatives:

$
\frac{dq}{dt}=2 \left[-\frac{1}{2}e^{-t/2}\cos(2t)-2e^{-t/2}\sin(2t)\right]
$

simplifying:

$
\frac{dq}{dt}=-e^{-t/2}[\cos(2t)+4\sin(2t)]
$

RonL
• Apr 29th 2006, 10:48 AM
CaptainBlack
Quote:

Hey guys,

Finding hard to solve these derivative:

1) $q = 2e^{-t/2}cos2t$ (don't know how to deal with the index -t/2)

2) $x = e^{2t}t^3(2-t)^4$ (this seem confusing as don't know what rule to apply)

3) $e^{x+y} + sin3x = 0$ (not to sure about index agian)

I hope someone can help. Thanks

What you seem to be having a problem with in 1) and 3) is:

$
\frac{d}{dx}e^{f(x)}
$

By the chain rule this is:

$
\frac{d}{dx}e^{f(x)}=f'(x)\ \frac{d}{dx}e^{y}\left\vert_{f(x)}\frac{}{}=f'(x)\ e^{f(x)}
$

RonL
• Apr 30th 2006, 01:57 AM
re:
Thanks for the quick response and the general rule! :)

So for 2) is it product and chain together?
• Apr 30th 2006, 02:28 AM
CaptainBlack
Quote:

Thanks for the quick response and the general rule! :)

So for 2) is it product and chain together?

Yes, though you will have the product:

$
x = \{e^{2t}\}\ \{t^3(2-t)^4\}
$

so you may need to use the product rule again for the derivative of the
second term on the RHS.

RonL
• Apr 30th 2006, 02:38 AM
CaptainBlack
Quote:

3) $e^{x+y} + sin3x = 0$ (not to sure about index agian)

This is an example of implicit differentiation (and the chain rule):

$
\frac{d}{dx} e^{x+y}+\frac{d}{dx}\sin(3x)=0
$
,

so:

$
e^{x+y}\ \frac{dy}{dx} + 3\cos(3x)=0\ \ \ \dots(1)
$

but from the original equation we know that:

$
e^{x+y}=-\sin(3x)
$
,

so substituting this into equation $(1)$ gives:

$
-\sin(3x)\frac{dy}{dx}+3\cos(3x)=0
$

or rearranging:

$
\frac{dy}{dx}=3\frac{\cos(3x)}{\sin(3x)}=3\cot(3x)
$

RonL
• Apr 30th 2006, 09:08 AM