# [SOLVED] more derivatives

• Oct 23rd 2008, 02:28 PM
iz1hp
[SOLVED] more derivatives
Find and simplify the following derivatives:

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

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

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

thanks :D
• Oct 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.
• Oct 23rd 2008, 02:52 PM
iz1hp
i know they are the chain rule which is

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

but I dont know how to do these chain rules with cos, sec and tan to exponents of more then 1
• Oct 23rd 2008, 03:04 PM
skeeter
$\displaystyle f(x) = [\sec(1 - \sqrt{4x})]^2$

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

do the cosine one.
• Oct 23rd 2008, 03:19 PM
iz1hp
Quote:

Originally Posted by skeeter
$\displaystyle f(x) = [\sec(1 - \sqrt{4x})]^2$

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

do the cosine one.

wouldnt this one be:

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

???
• Oct 23rd 2008, 03:28 PM
skeeter
$\displaystyle \frac{d}{dx} (-\sqrt{4x}) =$

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

$\displaystyle -2 \cdot \frac{d}{dx} (\sqrt{x}) = -2 \cdot \frac{1}{2\sqrt{x}} = -\frac{1}{\sqrt{x}}$
• Oct 23rd 2008, 03:34 PM
iz1hp
so the cos one would be

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

Originally Posted by iz1hp
so the cos one would be

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

almost ...

$\displaystyle f'(x) = 3[cos(3x^2-2)]^2 \cdot [-sin(3x^2-2)] \cdot (6x)$
• Oct 23rd 2008, 03:45 PM
iz1hp
so the tan one is...

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

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

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

??
• Oct 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.
• Oct 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

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

so far ive gotten to:

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

and now im stuck
• Oct 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

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

so far ive gotten to:

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

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