1. Find derivative and simplify

y=csc(x^2)
Ok on this one I know you use the chain rule? The formula for csc is -cscu cotu du/dx so should it go somthing like
------>y'= -csc(x^2) cot(x^2) (2x) ???

also on
f(x)=tan^3x
can I rewrite it to
-----> f(x)= (tan x)^3
f '(x)= 3(sec^2 x)^2 (1)

Any help would be great.

2. Originally Posted by Nightasylum
y=csc(x^2)
Ok on this one I know you use the chain rule? The formula for csc is -cscu cotu du/dx so should it go somthing like
------>y'= -csc(x^2) cot(x^2) (2x) ???

also on
f(x)=tan^3x
can I rewrite it to
-----> f(x)= (tan x)^3
f '(x)= 3(sec^2 x)^2 (1)

Any help would be great.
First one is right.

If $f(x)=(\tan(x))^3$ then $f'=3(\tan(x))^2*\sec^2(x)$. Make sense?

3. f(x)=tan^3x
can I rewrite it to
-----> f(x)= (tan x)^3
It is fair play for me to rewrite like that right?
Im still a little lost on the 2nd one.

4. Originally Posted by Nightasylum
f(x)=tan^3x
can I rewrite it to
-----> f(x)= (tan x)^3
It is fair play for me to rewrite like that right?
Im still a little lost on the 2nd one.
Yes you can write it like that. The tan^3(x) notation is just used for convenience and less parentheses, but both expressions are the same.

What if you had to take the derivative of u^3 and u is a function of x. The derivative would be 3x^2*du/dx. This is exactly what I did. The first step is the power rule then the second part is the chain rule.