Derivative of 1/cosx^2

David98678

New member
Joined
Feb 13, 2019
Messages
1
In my book it says cos^2 not x^2 to avoid confusion. If someone could explain how they solve it i would appreciate it.
 

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
3,656
In my book it says cos^2 not x^2 to avoid confusion. If someone could explain how they solve it i would appreciate it.
I do not understand why you would write cosx2 and then say it really is cos2x. Very strange.

Here is the rule you must use. Suppose u is anything. Then \(\displaystyle \frac{d}{dx}\)(u)n=n(u)n-1\(\displaystyle \frac{d}{dx}\)(u)

Please note that in this example, u is simply cos(x) and n= -2.
 
Last edited:

pka

Elite Member
Joined
Jan 29, 2005
Messages
8,326
In my book it says cos^2 not x^2 to avoid confusion. If someone could explain how they solve it i would appreciate it.
The notation \(\displaystyle \cos\) stands for the cosine function therefore we should use function notation.
That is use parentheses, \(\displaystyle \large\cos(x)\). Even though historically the notation \(\displaystyle \cos x\) was used, in practice we have move away from that.
If we have a function \(\displaystyle f(x)\) you surely know that there is a difference in \(\displaystyle f(x^2)~\&~f(x)^2\).
So that \(\displaystyle \cos^2(x)=\cos(x)^2\) BUT \(\displaystyle \cos^2(x)\ne\cos(x^2)\ne\cos(x)^2\).
If you still have confusion on this notation please post your questions.
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
4,149
In my book it says cos^2 not x^2 to avoid confusion. If someone could explain how they solve it i would appreciate it.
So, the problem is to differentiate \(\displaystyle \dfrac{1}{\cos^2 x}\)?

I would write it as \(\displaystyle (\cos x)^{-2}\) and use the chain rule. That is, take \(\displaystyle u = \cos x\), differentiate \(\displaystyle u^{-2}\), and multiply that by the derivative of \(\displaystyle \cos x\).

What do you get?
 

HallsofIvy

Elite Member
Joined
Jan 27, 2012
Messages
5,045
The notation \(\displaystyle \cos\) stands for the cosine function therefore we should use function notation.
That is use parentheses, \(\displaystyle \large\cos(x)\). Even though historically the notation \(\displaystyle \cos x\) was used, in practice we have move away from that.
If we have a function \(\displaystyle f(x)\) you surely know that there is a difference in \(\displaystyle f(x^2)~\&~f(x)^2\).
So that \(\displaystyle \cos^2(x)=\cos(x)^2\) BUT \(\displaystyle \cos^2(x)\ne\cos(x^2)\ne\cos(x)^2\).
If you still have confusion on this notation please post your questions.
Okay, I give up! I know what \(\displaystyle \cos^2(x)\) means and what \(\displaystyle cos(x^2)\) means. But what in the world is \(\displaystyle cos(x)^2\) if it is not a sloppy way of writing one of the others?
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
8,326
Okay, I give up! I know what \(\displaystyle \cos^2(x)\) means and what \(\displaystyle cos(x^2)\) means. But what in the world is \(\displaystyle cos(x)^2\) if it is not a sloppy way of writing one of the others?
Come on Prof Halls, \(\displaystyle \cos(x^2)\) is the cosine of the square of x , in the same way any function \(\displaystyle f(x^2)\) means \(\displaystyle f\) evaluated for \(\displaystyle x^2\).
That is a good reason for using \(\displaystyle \cos(x)\) in stead of \(\displaystyle \cos~x\).
Many calculators, basic programming,and computer algebras require the notation \(\displaystyle \cos(x)^2\) (actually any fuction) to return the square of the function, i.e. the routine does not recognize \(\displaystyle f^2(x)\) for \(\displaystyle f(x)^2\).

BTW: The post you quoted carefully said \(\displaystyle \cos^2(x)=\cos(x)^2\) but \(\displaystyle \cos^2(x)\ne\cos(x^2)\).
 
Last edited:

Otis

Senior Member
Joined
Apr 22, 2015
Messages
1,648

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
3,656

lookagain

Senior Member
Joined
Aug 22, 2010
Messages
2,428
cos(x)2 is the number cos(x) being squared
The former is poor form if it is intended to be equivalent for the square of the cosine of x, because it has ambiguous style. It should not be used.

Extra grouping symbols should be used for it instead, such as:

\(\displaystyle [cos(x)]^2\)
 
Last edited:

Otis

Senior Member
Joined
Apr 22, 2015
Messages
1,648
… it has ambiguous style. It should not be used …
It's not ambiguous, for folks who understand function notation. Is your position that it shouldn't be used because you think a majority of people do not understand function notation? :cool:
 

Otis

Senior Member
Joined
Apr 22, 2015
Messages
1,648
So we can't calculate sin(cos(x))???
I don't understand your statement (followed by question marks). Do you have a value for x?
 

lookagain

Senior Member
Joined
Aug 22, 2010
Messages
2,428
It's not ambiguous, for folks who understand function notation. Is your position that it shouldn't be used because you think a majority of people do not understand function notation? :cool:
No, I claim it is not consistent function notation for the exponentiation
of trig functions to begin with.
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
4,149
The former is poor form if it is intended to be equivalent for the square of the cosine of x, because it has ambiguous style. It should not be used.

Extra grouping symbols should be used for it instead, such as:

\(\displaystyle [cos(x)]^2\)
I agree that \(\displaystyle \cos(x)^2\) is at least a little ambiguous, and therefore risky to use, especially with inexperienced readers.

In a context where one always uses parentheses around function arguments, it is not ambiguous. So programmers will have no doubt as to what is intended.

But in trig and log problems in textbooks, it is still common to use the old, pre-function notation where \(\displaystyle \cos x\) or \(\displaystyle \log x\) is acceptable. And in that context, consider a slightly worse case: \(\displaystyle \cos(x+1)^2\). Here, the parentheses might be present to mark the argument, \(\displaystyle x+1\); or they might be there to distinguish the base of the power, the whole thing being the argument.

In an ideal world, a little thought may make it clear that the parentheses are intended to hold the argument; but expecting everyone to think carefully is "blaming the victim". Consideration for the reader demands that we remove stumbling blocks in case they are not accustomed to seeing things the same way we do. That's how good communication works: the transmitter plans the content to prevent errors on the part of the receiver, rather than putting the entire burden on the receiver.
 
Top