Unable to differentiate $\cos(x) \cos(2x) \cos(3x)$ and $\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$ I apologize for the lack of LaTeX. I will update this question with the proper LaTeX as soon as possible.
I am having trouble with two differentiation exercise questions and was hoping someone could provide a hint in the right direction.
The two questions are: Differentiate
$$\cos(x) \cos(2x) \cos(3x)$$
and
$$\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$$
Any hint in the right direction is highly appreciated.
Note: Please try to provide the hint towards a logarithmic approach as this was in the logarithms exercise. Also please provide only a hint, not the full answer. I'm hoping to savor some of the glory of mastering a half-tamed beast myself.
 A: Hint
Suppose that you have $$A=f(x)\times g(x)\times h(x)$$ So, using logarithms, you have $$\log(A)=\log(f(x))+\log(g(x))+\log(h(x))$$ So, taking derivatives $$\frac{1}{A}\frac{dA}{dx}=\frac{f'(x)}{f(x)}+\frac{g'(x)}{g(x)}+\frac{h'(x)}{h(x)}$$
I am sure you can take from here and apply the above to your two problems.
A: I hope you mean in the first case $\cos(x)\cdot \cos(2\cdot x)\cdot \cos(3\cdot x)$. Then you need the product rule here. And in the second case first the chain rule for differentiating $f(g(x))$ and then the quotient rule.
Hope that helps. I see IMHO no connection with logarithms, maybe someone else.
A: When you want to find the derivative of a product of 3 functions, you may regroup them thus: (f*g)*h and then consider that you have to find the derivative of a product of 2 functions: the first being f*g, the second being h: ( (f*g) * h)' = (f*g)'*h+(f*g)*h'. Then you derive (f*g)'=f'*g+f*g' giving (f*g*h)'=(f'*g+f*g')*h+(f*g)*h = f'gh+fg'h+fgh'.
As you see the result can be written: derive one function multiply it by the other functions and add all such products.
As an exercise, you might prove that this idea works for the derivative of the product of 4 functions.
