1. ## Deriving Sinosoid Functions

Could someone be so kind as to derive these functions for me? My answers are not what the back of the book has but i dont now where I go wrong so if you could show all your work I would appreciate it.

$y = cos^3x$

$y = sin(x^3)$

$f(x) = cos(5x-3)$

$f(x) = sin^2xcos (\frac{x}{2})$

$f(x) = cos^2(4x^2)$

$g(x) = \frac{cosx}{cosx-sinx}$

Thanks
Chris

2. Originally Posted by ffezz
Could someone be so kind as to derive these functions for me? My answers are not what the back of the book has but i dont now where I go wrong so if you could show all your work I would appreciate it.

$y = cos^3x$

$y = sin(x^3)$

$f(x) = cos(5x-3)$

$f(x) = sin^2xcos (\frac{x}{2})$

$f(x) = cos^2(4x^2)$

$g(x) = \frac{cosx}{cosx-sinx}$

Thanks
Chris
What do you mean derive? Differentiate these sinusoidal functions?

3. I think it's just about contemplating the derivative of each one.

4. Hi,

Originally Posted by ffezz
Could someone be so kind as to derive these functions for me? My answers are not what the back of the book has but i dont now where I go wrong so if you could show all your work I would appreciate it.
It's better for you to show your working so that we can more easily go through your mistakes and explain what's wrong

5. ## ok heres my work

yes i did mean differentiate =)

$f(x) = cos^2(4x^2)$

$f'(x) = 2cos(4x^2)-sin(4x^2)(8x)$

$= -16xcos(4x^2)-sin(4x^2)$
Thats as far as I can get and this appears to be a f(g(h(x))) function. The answer is cut off for this one so I dont know if its right or no

now this one I have no idea

$sin^2xcos(\frac{x}{2})$

$f'(x) = 2sinxcosxcos(\frac{x}{2})sin^2x-sin(\frac{x}{2})$

I dont know where to go from there

I just moved on to the next question and it is even more difficult.

"Dtetermine the equation of the tangent to

$y = 2 + cos2x$ at $x = \frac{5pi}{6}$

the answer is appearntly $\sqrt{3x} + 3 - \frac{5\sqrt{3}}{6}pi$

6. Ok, that's far better !

One thing, essential thing you have to keep in mind for this exercise is the chain rule :

$[f(g(x))]'=g'(x) \cdot f'(g(x))$

From this rule, you can do funny things, especially this one :

$[f(g(h(x)))]'=[f(k(x))]'$, with $k(x)=g(h(x))$

Using the chain rule :

$[f(k(x))]'=k'(x) \cdot f'(k(x))$

Now, what is $k'(x)$ ? $k'(x)=[g(h(x))]'=h'(x) \cdot g'(h(x))$

So $\boxed{[f(g(h(x)))]'=h'(x) \cdot g'(h(x)) \cdot f'(g(h(x)))}$

Originally Posted by ffezz
yes i did mean differentiate =)

$f(x) = cos^2(4x^2)$

$f'(x) = 2cos(4x^2)-sin(4x^2)(8x)$

$= -16xcos(4x^2)-sin(4x^2)$
Thats as far as I can get and this appears to be a f(g(h(x))) function. The answer is cut off for this one so I dont know if its right or no

now this one I have no idea

$sin^2xcos(\frac{x}{2})$

$f'(x) = 2sinxcosxcos(\frac{x}{2})sin^2x-sin(\frac{x}{2})$

I dont know where to go from there

I just moved on to the next question and it is even more difficult.

"Dtetermine the equation of the tangent to

$y = 2 + cos2x$ at $x = \frac{5pi}{6}$

the answer is appearntly $\sqrt{3x} + 3 - \frac{5\sqrt{3}}{6}pi$