Derivative qus.

**dadon** · April 29th 2006, 09:22 AM

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

dadon

**CaptainBlack** · April 29th 2006, 09:37 AM

Originally Posted by dadon

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

Start with the product rule:

$ \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

**CaptainBlack** · April 29th 2006, 09:48 AM

Originally Posted by dadon

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

dadon

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

**dadon** · April 30th 2006, 12:57 AM

Thanks for the quick response and the general rule!

So for 2) is it product and chain together?

**CaptainBlack** · April 30th 2006, 01:28 AM

Originally Posted by dadon

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

**CaptainBlack** · April 30th 2006, 01:38 AM

Originally Posted by dadon

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

**dadon** · April 30th 2006, 08:08 AM

Hi Thanks again!

Both methods should give the same answer?

**dadon** · April 30th 2006, 08:23 AM

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!

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