# Trigonometric Derivatives

• Nov 29th 2008, 10:01 PM
meiyukichan
Trigonometric Derivatives
I have a problem with understanding derivatives.
How do I find the derivative and also simplify it?

$\displaystyle x cos x$

$\displaystyle sec^2 3x$

An explanation would really help for me to understand better.
Thank you!

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I also need help on these derivative problems: CLICK
• Nov 30th 2008, 01:17 AM
earboth
Quote:

Originally Posted by meiyukichan
I have a problem with understanding derivatives.
How do I find the derivative and also simplify it?

$\displaystyle x cos x$

...

You are dealing with a product of 2 functions:

$\displaystyle f(x)=x \cdot \cos(x)$ ....... Use product rule to calculate the drivation:

If $\displaystyle f(x)=g(x) \cdot h(x)$ ....... then ....... $\displaystyle f'(x)=h(x) \cdot g'(x) + g(x) \cdot h'(x)$

Therefore:

$\displaystyle f(x)=x \cdot \cos(x)~\implies~f'(x)=\cos(x) \cdot 1 + x\cdot (-\sin(x))~\implies~f'(x)=\cos(x)-x\sin(x)$
• Nov 30th 2008, 01:17 AM
Mathstud28
Quote:

Originally Posted by meiyukichan
I have a problem with understanding derivatives.
How do I find the derivative and also simplify it?

$\displaystyle x cos x$

$\displaystyle sec^2 3x$

An explanation would really help for me to understand better.
Thank you!

--------
I also need help on these derivative problems: CLICK

You need four facts, namely:

$\displaystyle \frac{d}{dx}\bigg[f(g(x))\bigg]=f'(g(x))\cdot{g'(x)}$

$\displaystyle \frac{d}{dx}\bigg[f(x)\cdot{g(x)}\bigg]=f'(x)g(x)+g'(x)f(x)$

$\displaystyle \frac{d}{dx}\bigg[\cos(x)\bigg]=-\sin(x)$

$\displaystyle \frac{d}{dx}\bigg[\sec(x)\bigg]=\sec(x)\tan(x)$

Now use those in combination to attaing the desired derivatives.
• Nov 30th 2008, 09:47 AM
meiyukichan
Quote:

Originally Posted by Mathstud28
You need four facts, namely:

$\displaystyle \frac{d}{dx}\bigg[f(g(x))\bigg]=f'(g(x))\cdot{g'(x)}$

$\displaystyle \frac{d}{dx}\bigg[f(x)\cdot{g(x)}\bigg]=f'(x)g(x)+g'(x)f(x)$

$\displaystyle \frac{d}{dx}\bigg[\cos(x)\bigg]=-\sin(x)$

$\displaystyle \frac{d}{dx}\bigg[\sec(x)\bigg]=\sec(x)\tan(x)$

Now use those in combination to attaing the desired derivatives.

I don't understand what you mean by that.
Can you show me how in an example similar to $\displaystyle sec^2 3x$
or you can just use that as an example because it'll be easy for me to understand for similar problems.
• Nov 30th 2008, 09:57 AM
r_maths
Quote:

Originally Posted by meiyukichan
I don't understand what you mean by that.
Can you show me how in an example similar to $\displaystyle sec^2 3x$
or you can just use that as an example because it'll be easy for me to understand for similar problems.

This is a great website for finding derivatives: http://www.mathhelpforum.com/math-he...step-step.html

The integration version is just as good for checking answers but it doesn't give you step-by-step solutions.
• Nov 30th 2008, 12:17 PM
meiyukichan
Quote:

Originally Posted by r_maths
This is a great website for finding derivatives: http://www.mathhelpforum.com/math-he...step-step.html

The integration version is just as good for checking answers but it doesn't give you step-by-step solutions.

Okay. Thank you for the advice/tip/help! :D
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