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So, I once read somewhere that if: $y = \tan x, \quad\frac{dy}{dx}=\sec^2x$.

I think I have found the proof, but I am not so sure.


Recall the identity:

$$\tan x \equiv \frac{\sin x}{\cos x}$$

Recall the quotient rule:

(where $u$ and $v$ are each functions:) $$y=\frac{u}{v}, \quad \frac{dy}{dx}=\frac{v * \frac{du}{dx}-u* \frac{dv}{dx}}{v^2}$$

I did:

$u=\sin x$

$v = \cos x$

$\frac{du}{dx} = \cos x$

$\frac{dv}{dx} = -\sin x$

I believe that if you substitute all these values in, you get:

$$\frac{\cos^2x + \sin^2x}{\cos^2x}$$

Recall the identity:

$$\sin^2x + \cos^2x \equiv 1$$

This is the case on the top so:

$$\frac{dy}{dx}=\frac{1}{\cos^2x}$$


On the same webpage that I found these identities in, I also found that:

$$\sec^2x \equiv\frac{1}{\cos^2x}$$

This means that $$\frac{dy}{dx}=\sec^2x \equiv\frac{1}{\cos^2x}$$

But why does $\frac{1}{\cos^2x} \equiv \sec^2x $? I want a proof that the $\sec \leftarrow \rightarrow \cos$ identity works. From my research, not much has touched on this, so how is it provable that $\frac{dy}{dx}$ of $\tan x = \sec x^2$?

P.S. I have a hunch that it is something that is taken for granted, and that we don't necessarily have to know this.

P.P.S. Could you check this proof for me, and state if I have missed any steps?

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    $\begingroup$ This is true by definition: $$\sec{x}\equiv \frac{1}{\cos{x}}$$ $\endgroup$ – projectilemotion May 4 '17 at 15:37
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    $\begingroup$ The definition of $\sec$ is $\frac1\cos$. $\endgroup$ – Arthur May 4 '17 at 15:38
  • $\begingroup$ @projectilemotion , I know, but is there a proof? $\endgroup$ – VortexYT May 4 '17 at 15:38
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    $\begingroup$ @simplest_mathematics $\sec(x) = \frac{1}{\cos(x)}$ by definition -- you may be interested to read math.stackexchange.com/questions/1399781/… $\endgroup$ – Marcus Andrews May 4 '17 at 15:39
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    $\begingroup$ @simplest_mathematics, $\sec(x)$ just means $\frac{1}{\cos(x)}$, so it's not something you can prove, that's just what it means. Basically at some point in history people just decided it was inconvenient to always be writing $\frac{1}{\cos(x)}$ so they gave it a new name. $\endgroup$ – DMcMor May 4 '17 at 15:40
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But why does $\frac{1}{\cos^2x} \equiv \sec^2x $? I want a proof that the $\sec \leftarrow \rightarrow \cos$ identity works.

There is nothing to prove because this is a definition. We simply give another name to $\tfrac{1}{\cos x}$, we call it $\sec x$. You can avoid using $\sec x$ by replacing every occurence by $\tfrac{1}{\cos x}$, if you prefer. Using it allows to rewrite some formulas in a shorter and/or more elegant way.

Compare it to giving the name $\tan x$ to the quotient $\tfrac{\sin x}{\cos x}$: you can't prove this either, at least not if this is how you define $\tan x$.

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$\sec x = \frac 1{\cos x}$ by definition.

However, there is a nice geometric proof that $\frac {d}{dx} \tan x = \sec^2 x$ that is worth knowing. enter image description here

In the figure we have two right triangles. The base is 1.

Lets call the angle at the center of the unit circle $x, h, x+h$

The heights of the right triangles are $\tan x$ and $\tan (x+h)$ hypoteni are $\sec x$ and $\sec (x+h)$

The areas are $\frac 12 \tan x$ and $\frac 12 \tan (x+h)$

The small triangle with angle $h$ has area $\frac 12 \tan (x+h) - \frac 12 \tan x$

The area of this triangle is less than the area of the section of the circle with radius $\sec (x+h)$ and greater than the section of the circle of radius $\sec x$

These areas are $\frac 12 (\sec^2 (x+h)) h$ and $\frac 12 (\sec^2 x) h$

So,

$\frac 12 (\sec^2 x) h \le \frac 12 (\tan (x+h) - \tan x) \le \frac 12 (\sec^2 (x+h)) h$

or

$(\sec^2 x)\le \frac {\tan (x+h) - \tan x}{h} \le \sec^2 (x+h)$

Now take the limit as h goes to 0.

$\frac {d}{dx} \tan x = \sec^2 x$

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