# Derivative qus.

• Apr 29th 2006, 09:22 AM
Derivative qus.
Hey guys,

Finding hard to solve these derivative:

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

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

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

I hope someone can help. Thanks

• Apr 29th 2006, 09:37 AM
CaptainBlack
Quote:

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

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

so using the chain rule to do the derivatives:

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

simplifying:

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

RonL
• Apr 29th 2006, 09:48 AM
CaptainBlack
Quote:

Hey guys,

Finding hard to solve these derivative:

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

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

3) $\displaystyle e^{x+y} + sin3x = 0$ (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:

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

By the chain rule this is:

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

RonL
• Apr 30th 2006, 12:57 AM
re:
Thanks for the quick response and the general rule! :)

So for 2) is it product and chain together?
• Apr 30th 2006, 01:28 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:

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

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

RonL
• Apr 30th 2006, 01:38 AM
CaptainBlack
Quote:

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

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

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

so:

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

but from the original equation we know that:

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

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

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

or rearranging:

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

RonL
• Apr 30th 2006, 08:08 AM