[SOLVED] more derivatives

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October 23rd 2008, 02:28 PM
iz1hp

[SOLVED] more derivatives

Find and simplify the following derivatives:

$ f(x) = cos^3 (3x^2 - 2) $

$ f(x) = sec^2 (1 - \sqrt{4x}) $

$ f(x) = tan^3 (\sqrt{1-x^3}) $

thanks :D
October 23rd 2008, 02:41 PM
skeeter

all of these are derivatives involving the use the chain rule ... let's see how you do at least one of them.
October 23rd 2008, 02:52 PM
iz1hp

i know they are the chain rule which is

$ g'(h(x)) \cdot h'(x) $

but I dont know how to do these chain rules with cos, sec and tan to exponents of more then 1
October 23rd 2008, 03:04 PM
skeeter

$f(x) = [\sec(1 - \sqrt{4x})]^2$

$f'(x) = 2\sec(1 - \sqrt{4x}) \cdot \sec(1 - \sqrt{4x}) \cdot \tan(1 - \sqrt{4x}) \cdot \left(-\frac{1}{\sqrt{x}}\right)$

do the cosine one.
October 23rd 2008, 03:19 PM
iz1hp

Quote:

Originally Posted by skeeter

$f(x) = [\sec(1 - \sqrt{4x})]^2$

$f'(x) = 2\sec(1 - \sqrt{4x}) \cdot \sec(1 - \sqrt{4x}) \cdot \tan(1 - \sqrt{4x}) \cdot \left(-\frac{1}{\sqrt{x}}\right)$

do the cosine one.

wouldnt this one be:

$ f'(x) = 2sec(1 - \sqrt{4x}) \cdot sec(1 - \sqrt{4x}) \cdot tan(1 - \sqrt{4x}) \cdot (\frac{-1}{2\sqrt{4x}}) $

???
October 23rd 2008, 03:28 PM
skeeter

$\frac{d}{dx} (-\sqrt{4x}) =$

$\frac{d}{dx} (-2\sqrt{x}) =$

$-2 \cdot \frac{d}{dx} (\sqrt{x}) = -2 \cdot \frac{1}{2\sqrt{x}} = -\frac{1}{\sqrt{x}}$
October 23rd 2008, 03:34 PM
iz1hp

so the cos one would be

$ f'(x) = 3[cos(3x^2-2)]^2 \cdot [cos(3x^2-2)] \cdot [-sin(3x^2-2)] \cdot (6x) $
October 23rd 2008, 03:36 PM
skeeter

Quote:

Originally Posted by iz1hp

so the cos one would be

$ f'(x) = 3[cos(3x^2-2)]^2 \cdot [cos(3x^2-2)] \cdot [-sin(3x^2-2)] \cdot (6x) $

almost ...

$f'(x) = 3[cos(3x^2-2)]^2 \cdot [-sin(3x^2-2)] \cdot (6x)$
October 23rd 2008, 03:45 PM
iz1hp

so the tan one is...

$ f(x) = tan^3(\sqrt{1-x^3}) $

$ f(x) = [tan(\sqrt{1-x^3})]^3 $

$ f'(x) = 3[tan(\sqrt{1-x^3})]^2 \cdot [sec^2(\sqrt{1-x^3})] \cdot (\frac{1}{2\sqrt{1-x^3}}) \cdot (-3x^2) $

??
October 23rd 2008, 03:52 PM
skeeter

... and you said you didn't know how.

(Clapping)

good job. now clean up the algebra on all of them.
October 23rd 2008, 03:59 PM
iz1hp

ahhah thanks for explaining it to me haha :P

if you can help me with one more, ill be set :P

$ f(x) = cos(sin(cos x)) $

so far ive gotten to:

$ f'(x) = -sin(sin(cos x)) \cdot cos(cos x) \cdot -sin x $

and now im stuck
October 23rd 2008, 04:04 PM
Jameson

Quote:

Originally Posted by iz1hp

ahhah thanks for explaining it to me haha :P

if you can help me with one more, ill be set :P

$ f(x) = cos(sin(cos x)) $

so far ive gotten to:

$ f'(x) = -sin(sin(cos x)) \cdot cos(cos x) \cdot -sin x $

and now im stuck

Hey iz1hp.

How about starting a new thread for this new question? It's a habit I'd like you and all users to get into. I'm going to close this thread but please make a new one!

Jameson