1. ## 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

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

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

4. ## re:

Thanks for the quick response and the general rule!

So for 2) is it product and chain together?

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

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

7. Hi Thanks again!

Both methods should give the same answer?

8. The implicit way is a much neater, when applying the product and chain rule it gets abit messy that's what I have been trying!