Finding the derivative of cosx from first principles

[BOAS]

Hello,

i'm confused by a certain step in my working for finding the derivative of cosx, I don't know for sure that it's right, but I'm following the same approach my book uses to show the derivative of sinx.

Homework Statement

Find \frac{d}{dx}cosx from first principles.

Homework Equations

The Attempt at a Solution

I don't know how to show my limits in LaTeX so please excuse that. They are shown in my written working.

\frac{d}{dx}f(x) = \lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x}

If f(x) = cosx

\frac{d}{dx}cosx = \lim_{\delta x \to 0} \frac{cos(x + \delta x) - cosx}{\delta x}

cos(x + \delta x) - cosx \equiv -2sin(x + \frac{\delta x}{2})sin\frac{\delta x}{2} (the factor formulae)

\frac{d}{dx}cosx = \lim_{\delta x \to 0} -sin(x + \frac{\delta x}{2}) \frac{sin\frac{\delta x}{2}}{\frac{\delta x}{2}}

It is that last step that I don't follow entirely. Frankly, I did it because a similar thing is done in my textbooks derivation of d/dx sinx. Specifically, I don't follow what they have done with the 2 that was in front of the trig identity. The bottom of the equation looks weird to me.

I go on to say (again the limits are shown in my written working)

\lim_{\delta x \to 0} [-sin(x + \frac{\delta x}{2})] = -sinx

\lim_{\delta x \to 0} [\frac{sin\frac{ \delta x}{2}}{\frac{\delta x}{2}}] = 1 (i'm also a little unsure of why this last bit is equal to 1).

\frac{d}{dx}cosx = -sinx

I hope the bits that are confusing me are clear...

Thanks.

EDIT - To add in the limits

 

[jedishrfu]

multiplying by 2 is the same as dividing by 1/2

ie to refactor the 2 multiply by the factor (1/2) / (1/2) so the two is canceled out and a 1/2 is left in the denominator of the fraction.

 

[BOAS]

So what's going on there is this;

cos(x + \delta x) - cosx \equiv -2sin(x + \frac{\delta x}{2})sin\frac{\delta x}{2}

\frac{-2sin(x + \frac{\delta x}{2})sin\frac{\delta x}{2}}{\delta x}* \frac{1/2}{1/2} = \frac{-sin(x + \frac{\delta x}{2})sin\frac{\delta x}{2}}{\delta x *1/2} = \frac{-sin(x + \frac{\delta x}{2})sin\frac{\delta x}{2}}{\frac{\delta x}{2}}

\frac{d}{dx}cosx = -sin(x + \frac{\delta x}{2}) \frac{sin\frac{\delta x}{2}}{\frac{\delta x}{2}}

I think that makes sense.

Thanks.

 

[Ray Vickson]

So what's going on there is this;

cos(x + \delta x) - cosx \equiv -2sin(x + \frac{\delta x}{2})sin\frac{\delta x}{2}

\frac{-2sin(x + \frac{\delta x}{2})sin\frac{\delta x}{2}}{\delta x}* \frac{1/2}{1/2} = \frac{-sin(x + \frac{\delta x}{2})sin\frac{\delta x}{2}}{\delta x *1/2} = \frac{-sin(x + \frac{\deltqa x}{2})sin\frac{\delta x}{2}}{\frac{\delta x}{2}}

\frac{d}{dx}cosx = -sin(x + \frac{\delta x}{2}) \frac{sin\frac{\delta x}{2}}{\frac{\delta x}{2}}

I think that makes sense.

Thanks.

What you wrote is not correct:
\frac{d}{dx} \cos(x) \neq -\sin(x + \delta x /2) <br /> \frac{\sin(\delta x /2)}{\delta x /2},
because
\frac{d}{dx} \cos(x) \neq \frac{\cos(x + \delta x) - \cos(x)}{\delta x}
What is true is
\frac{d}{dx} \cos(x) = \lim_{\delta x \to 0} \frac{\cos(x + \delta x) - \cos(x)}{\delta x}
In other words, you need to take a limit.

 

[BOAS]

What you wrote is not correct:
\frac{d}{dx} \cos(x) \neq -\sin(x + \delta x /2) <br /> \frac{\sin(\delta x /2)}{\delta x /2},
because
\frac{d}{dx} \cos(x) \neq \frac{\cos(x + \delta x) - \cos(x)}{\delta x}
What is true is
\frac{d}{dx} \cos(x) = \lim_{\delta x \to 0} \frac{\cos(x + \delta x) - \cos(x)}{\delta x}
In other words, you need to take a limit.

If you read my op again you'll notice I apologise for not including limits because I didn't know how to write them in LaTeX, but they are included in my written work.

Looking at your post, I do now know how to write them though :)

With that in mind, is my work still 'wrong'?

EDIT - I'll go back and put them in the relevant places

 

[maajdl]

"i'm also a little unsure of why this last bit is equal to 1"

Look on a circle: for small angles the "opposite segement" is almost the angle.
See:
http://en.wikipedia.org/wiki/Small-angle_approximation
http://en.wikipedia.org/wiki/File:Small_angle_triangle.svg

 

[BOAS]

"i'm also a little unsure of why this last bit is equal to 1"

Look on a circle: for small angles the "opposite segement" is almost the angle.
See:
http://en.wikipedia.org/wiki/Small-angle_approximation
http://en.wikipedia.org/wiki/File:Small_angle_triangle.svg

Does this imply that

\lim_{\delta x \to 0} dx/dx = 1?

 

[Ray Vickson]

If you read my op again you'll notice I apologise for not including limits because I didn't know how to write them in LaTeX, but they are included in my written work.

Looking at your post, I do now know how to write them though :)

With that in mind, is my work still 'wrong'?

EDIT - I'll go back and put them in the relevant places

Your basic work was OK, but instructors usually frown upon the use of an '=' sign between things that are not equal; doing that is a good way to lose marks for no good reason.

To get a limit expression in LaTeX, use "\ lim_{h \to 0} " (with no space between the '\' and the word 'lim').If you don't use a backslash it won't look very good---just like your 'sin' and 'cos' do not look good. Here is what I mean:
lim_{h \to 0} sin(h)/h \text{ using ``lim_{h \to 0} sin(h)/h&#039;&#039; } \\<br /> \\<br /> \lim_{h \to 0} \sin(h)/h \text{ using ``\lim&#039;&#039; and ``\sin&#039;&#039; }

 

[SteamKing]

hello,

i'm confused by a certain step in my working for finding the derivative of cosx, i don't know for sure that it's right, but I'm following the same approach my book uses to show the derivative of sinx.

Homework Statement

find \frac{d}{dx}cosx from first principles.

Homework Equations

The Attempt at a Solution

i don't know how to show my limits in latex so please excuse that. They are shown in my written working.

\frac{d}{dx}f(x) = \lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x}

if f(x) = cosx

\frac{d}{dx}cosx = \lim_{\delta x \to 0} \frac{cos(x + \delta x) - cosx}{\delta x}

I think you are OK to this point. I suggest you expand cos(x+δx) using the identity for the cosine of the sum of two angles, then substitute and take the limit as δx goes to zero.

 

[HallsofIvy]

How you show that \lim_{x\to 0} \frac{sin(x)}{x}=1 depends strongly upon how you have defined sine and cosine to begin with! Maajdl's suggestion, defining sine and cosine on a unit circle is most common in Calculus texts but it is also possible to define sine and cosine in terms of power series or as solutions to a differential equation.

 

[vanhees71]

But it's
\lim_{x \rightarrow 0} \frac{\sin x}{x}=1.

 

[HallsofIvy]

Yes, thanks. I have corrected that.

 

[SteamKing]

Much simpler to use the addition identity for the cosine.

 

[vanhees71]

I think, it's easier to define all elementary functions via power series and then their inverse functions. The geometrical meaning of functions like the trigonometric functions and the hyperbolic functions then is easily treated with analytical geometry in Euclidean \mathbb{R}^2. You can also easily prove the addition identities etc. with help of the series expansions.

 

[SteamKing]

That may be true if you are trying to derive all of mathematics from a few basic definitions. But all the OP needs to do is show how the derivative of the cosine may be derived from the basic definition. The cosine addition identity can be established with simple geometry. There is no need to prove the limit of sin(x)/x as x goes to zero, etc. Like Occam said, why eat the whole banquet when all you want is a donut?

 

[epenguin]

IMHO too much of an unnecessary big meal is made of the proof of this (which in my day we did at school not uni) and all anyone needs for proof is here: https://www.physicsforums.com/showpost.php?p=3670158&postcount=6 Just geometrical. I don't say some of the other things mentioned, e.g. the series, are not valuable, but IMHO they are valuable additionally, get to them from this not vice versa.

 

[BOAS]

Thank you for all the responses/discussion - apologies for the late reply.

I understand the proof now - I don't know if i'll ever need to use it, but it's nice to know.

@epenguin I am working through an A-level textbook, not a university one. Although I am at university, I'm doing a foundation physics year since I did not take A-level maths and subsequently regretted it.

 

[vela]

There is no need to prove the limit of sin(x)/x as x goes to zero.

Just wanted to note that this particular result is commonly established when the concept of limits is reviewed as a precursor to introducing the concept of derivatives, so it's not unreasonable to assume students should know this limit when trying to find the derivative of cosine using the definition.

 

[SteamKing]

But it's also not necessary to know it if you use the identity for the cosine of the sum of two angles, either.

 

[vela]

Really? Because I see that limit show up twice if you do it that way. Maybe we're thinking of different identities.

 

[gopher_p]

IMHO too much of an unnecessary big meal is made of the proof of this (which in my day we did at school not uni) and all anyone needs for proof is here: https://www.physicsforums.com/showpost.php?p=3670158&postcount=6 Just geometrical. I don't say some of the other things mentioned, e.g. the series, are not valuable, but IMHO they are valuable additionally, get to them from this not vice versa.

Since $dx\neq|ab|$ in the picture, your "proof" is unconvincing.

Some fairly minor modifications to your argument would establish the desired results, so it's definitely salvageable. I would point out, though, that this is very similar to the standard argument that is used to demonstrate $\lim\limits_{x\rightarrow 0}\frac{\sin x}{x}=1$.

From a pedagogical standpoint, I'd say it's maybe not a terrible idea to reinforce geometry ideas in the proof of the sinc limit and then reinforce some trig identities, limit laws, and definition of the derivative in the proof of the derivative formulas.