How do I find the derivative of tan(x) from first principles?

[Dr-NiKoN]

$f(x) = \tan(x) = \frac{\sin(x)}{\cos(x)}$
Using: $(\Delta x = h)$

$f'(x) = \frac{f(x + h) - f(x)}{h}$

$f'(x) = \frac{\frac{\sin(x + h)}{\cos(x + h)} - \frac{sin(x)}{\cos(x)}}{h}$

Where do I go from here?

 

[Dr-NiKoN]

$\sin(x + h) = -\sin(x)$
$\cos(x + h) = -cos(x)$

$f'(x) = \frac{\frac{-\sin(x)}{-\cos(x)} - \frac{\sin(x)}{\cos(x)}}{h}$

Is this the correct step?
hmpf, probably not..

 

[Dr-NiKoN]

Maybe:

$sin(u + v) = \sin(u)\cos(v) + \cos(u)\sin(v)$

Thus:

$f'(x) = \frac{\frac{\sin(x)\cos(h) + \cos(x)\sin(h)}{\cos(x)\cos(h) - \sin(x)\sin(h)} - \frac{sin(x)}{\cos(x)}}{h}$

?

 

[Galileo]

First try finding the derivative of sin(x) and cos(x) first. (You probably know them already).
Then use the quotient rule.
$$\left(\frac{f}{g}\right)'=\frac{gf'-fg'}{g^2}$$

 

[Dr-NiKoN]

Yes, I have no problems doing it that way.
But, how about using the general formula for derivatives:

$f'(x) = \frac{f(x + h) - f(x)}{h}$

Possible?

 

[TenaliRaman]

definitely,

Needs a bit of trigonometry result ...
tan(A-B) = [tan(A) - tan(B)]/[1+tan(A)tan(B)]
therefore,
tan(A) - tan(B) = tan(A-B)*(1+tan(A)tan(B))

tan(x+h)-tan(x) = tan(h)*(1+tan(x+h)tan(x))
Can u finish off now?

-- AI

 

[Dr-NiKoN]

Hm..
$f'(x) = \frac{\tan(h) * (1 + \tan(x+h)\tan(x))}{h}$

I'm not sure how I would proceed from here?

 

[TenaliRaman]

i am sure u know,
$$f'(x) = \lim_{h->0} \frac{f(x+h)-f(x)}{h}$$

so what is lim_{h->0} tan(h)/h ??

-- AI

 

[Dr-NiKoN]

$\frac{tan(x)}{x} = \frac{\frac{\sin(x)}{\cos(x)}}{x} = \frac{\sin(x)}{x}\frac{1}{cos(x)} = 1$

I'm not seeing the bigger picture though :)

Hm.
$f'(x) = \frac{\tan(h)}{h} \frac{1 + \tan(x+h)\tan(x)}{1}$

Is this what you mean?

 

[devious_]

$$f'(x)=\lim_{h\,\rightarrow\,0}1+\tan(x+h)\tan(x)$$

What is $$\lim_{h\,\rightarrow\,0}\tan(x+h)$$?

 

[NateTG]

$$f'(x)=\lim_{h\,\rightarrow\,0}1+\tan(x+h)\tan(x)$$

What is $$\lim_{h\,\rightarrow\,0}\tan(x+h)$$?

Nitpicking: Well, you also have to insure that $$\tan(x)$$ is defined. If $$x=\frac{\pi}{2}$$ the limit does not exist or is asymptotic.

 

[Nylex]

Nitpicking: Well, you also have to insure that $$\tan(x)$$ is defined. If $$x=\frac{\pi}{2}$$ the limit does not exist or is asymptotic.

Nitpicking: you mean ensure :).

 

[Dr-NiKoN]

Hm, I'm not seeing how to get to 1/cos^2x, but it doesn't matter, I'll just use f'(x)/g'(x).

$f(x) = \tan(x) \rightarrow (\frac{\sin(x)}{\cos(x)})' = \frac{(\sin(x))'\cos(x) - \sin(x)(\cos(x))'}{(\cos(x))^{2}} = \frac{\cos^{2}(x) + \sin^{2}(x)}{\cos^{2}(x)}$

Here is another question though:
Find the derivative of arctan(x), using arctan(tan(x)) = x.
This also has me stumped.

$g'(x) = \arctan(x)$

I know that:
$f'(x) = f'[g(x)] * g'(x)$

But, I don't see what I can do with arctan(x), except for the obvious:

$\arctan(\arctan(\tan(x)))$
which doesn't help at all.

How would I proceed? I'm suppoed to use the derivative of a "functionsfunction", not sure what it's called in english.

 

[hedlund]

Hm, I'm not seeing how to get to 1/cos^2x, but it doesn't matter, I'll just use f'(x)/g'(x).

$f(x) = \tan(x) \rightarrow (\frac{\sin(x)}{\cos(x)})' = \frac{(\sin(x))'\cos(x) - \sin(x)(\cos(x))'}{(\cos(x))^{2}} = \frac{\cos^{2}(x) + \sin^{2}(x)}{\cos^{2}(x)}$

Here is another question though:
Find the derivative of arctan(x), using arctan(tan(x)) = x.
This also has me stumped.

$g'(x) = \arctan(x)$

I know that:
$f'(x) = f'[g(x)] * g'(x)$

But, I don't see what I can do with arctan(x), except for the obvious:

$\arctan(\arctan(\tan(x)))$
which doesn't help at all.

How would I proceed? I'm suppoed to use the derivative of a "functionsfunction", not sure what it's called in english.

Say y = arctan(x) then tan(y) = tan(atan(x)). But tan(atan(x))=x for all x. So we have tan(y) = x deriving with respect to x on left side and right side we get:
y' * (tan(y)^2 + 1) = 1 => y' = 1/(tan(y)^2+1). But what is tan(y)? Well tan(y)=x so the derivate of arctan(x) is 1/(x^2+1).

 

[MiGUi]

Mmm I know to do it but with complex function, since

$$\arctan{z} = {1 \over 2} i ( \ln(1 -iz) - \ln(1 + iz))$$

You easily obtain that

$${d \over dz} \arctan{z} = {1 \over 1 + z^2}$$

I don't know how to do in other way, without knowing the expression of arctan in more elemmental functions...

 

[Dr-NiKoN]

Say y = arctan(x) then tan(y) = tan(atan(x)). But tan(atan(x))=x for all x. So we have tan(y) = x deriving with respect to x on left side and right side we get:
y' * (tan(y)^2 + 1) = 1 => y' = 1/(tan(y)^2+1). But what is tan(y)? Well tan(y)=x so the derivate of arctan(x) is 1/(x^2+1).

Hm.
$y = \arctan(x)$
$\tan(y) = x$
$\arctan(\tan(x)) = x$
This I understand.
But, how do you get from:
$y = \arctan(x)$
to
$tan(y) = \tan(\arctan(x))$

Why isn't it, or wouldn't it be:
$tan(y) = \arctan(\tan(x))$
?

 

[NateTG]

Hm.

But, how do you get from:
$y = \arctan(x)$
to
$tan(y) = \tan(\arctan(x))$

?

Take the tangent of both sides of the equation.

 

[Dr-NiKoN]

Ah, of course :)
so
$\tan(y) = \tan(\arctan(x)) = x$
And:
$f'[g(x)] * g'(x)$

$f(x) = \tan(y)$ and $g(x) = y$

$(tan(y))' * y' = x'$

$\frac{1}{\cos^2(x)} * 1 = 1$

I'm not understanding this :(
How do you get $\frac{1}{\cos^2(x) + 1}$
?

 

[MiGUi]

> Take the tangent of both sides of the equation.

I don't know if that is too correct since the tangent is defined between [0, pi/2) so what if x = n*pi/2?

 

[NateTG]

Ah, of course :)
so
$\tan(y) = \tan(\arctan(x)) = x$
And:
$f'[g(x)] * g'(x)$

$f(x) = \tan(y)$ and $g(x) = y$

$(tan(y))' * y' = x'$

$\frac{1}{\cos^2(x)} * 1 = 1$

I'm not understanding this :(
How do you get $\frac{1}{\cos^2(x) + 1}$
?

$$\tan (\arctan (x))=x$$
$$\frac{d}{dx} \tan (\arctan (x)) = \frac{d}{dx} x$$
$$\arctan'(x) \times \sec^2(\arctan(x))=1$$
$$\arctan'(x) = \cos^2(\arctan(x))$$
Now
$$\cos(\arctan(x))=\frac{\pm 1}{\sqrt{1-x^2}}$$
(Consider, for example, a right triangle where the adjacent side is 1, the opposite side is $$x$$ and the hypotenuse is $$\sqrt{1+x^2}$$)
So we can substitute that in:
$$\arctan'(x)=(\frac{\pm 1}{\sqrt{1-x^2}})^2=\frac{1}{1+x^2}$$
the $$\pm$$ drops out because of the square.

 

[Dr-NiKoN]

What is d/dx and sec?

 

[NateTG]

> Take the tangent of both sides of the equation.

I don't know if that is too correct since the tangent is defined between [0, pi/2) so what if x = n*pi/2?

There's no real problem - $$\arctan(\frac{n\pi}{2})$$ is a real number. It might be interesting if $$\arctan(x) = \frac{n\pi}{2}$$ but I would be interested to see a suitable (real) value $$x$$.

P.S. Dr. NiKoN : The derivative rule for function composition is called the "chain rule" in English.

 

[NateTG]

What is d/dx and sec?

$$\frac{d}{dx}$$ is a common notation for derivatives. For now, you could just think of it as $$\frac{d}{dx} f(x) =f'(x)$$.

And $$\sec(x)=\frac{1}{\cos(x)}$$ (it's called the secant).

 

[Dr-NiKoN]

I'm getting more and more confused :(
How do you get from:
$(\arctan(x))' * \frac{1}{\cos^2(x)}\arctan(x) = 1$
to
$(\arctan(x))' = \cos^2(\arctan(x))$

I don't really understand the following steps either. What mathematically steps are you taking here?

 

[NateTG]

I'm getting more and more confused :(
How do you get from:
$(\arctan(x))' * \frac{1}{\cos^2(x)}\arctan(x) = 1$
to
$(\arctan(x))' = \cos^2(\arctan(x))$

I don't really understand the following steps either. What mathematically steps are you taking here?

No, I had
$$(\arctan(x))'\times \frac{1}{\cos^2(\arctan(x))}=1$$
so
$$(\arctan(x))'=\cos^2(\arctan(x))$$
It's the cosine (or secant) of the arctangent, not the product of the cosine or secant and the arctangent.

$$\cos(\arctan(x))=\frac{\pm 1}{\sqrt{1+x^2}}$$
Is a trig identity. You should be able to derive it by drawing a right triangle, and thinking about it.

 

[Dr-NiKoN]

I'll try drawing up a triangle and see if I understand it that way.
thanks :)

 

[devious_]

First question:
$$\lim_{h\to0}\tan(x+h)=\tan(x)$$
So:
$$f'(x)=\lim_{h\to0}1+\tan(x+h)\tan(x)=1+\tan^2(x)=\sec^2x$$

Second question:
$$y=\arctan(x)$$
$$\tan(y)=x$$
$$\frac{dx}{dy}=\sec^2(y)=1+\tan^2(y)\ \ [\tan^2(y)=x^2]$$
$$\frac{dx}{dy}=1+x^2$$
$$\frac{dy}{dx}=\frac{1}{1+x^2}$$

 

[JasonRox]

I think that the derivative of arctan is the most beautiful out of all the trig functions.

 

[Dr-NiKoN]

I don't! :(

So we have tan(y) = x deriving with respect to x on left side and right side we get:
y' * (tan(y)^2 + 1) = 1

$\tan(y) = x$

Using: f'(x) = f'[g(x)] * g'(x)

$(\tan(y))' * y' = x'$

$\frac{1}{\cos^2x} * y' = x'$

I still don't understand this :(

 

[Dr-NiKoN]

Hm
$\tan(y) = x$ so $y = \arctan(x)$

$(\tan(y))' * y' = x'$

I know what that $y = \arctan(x)$
So
$(\tan(\arctan(x))' * y' = x'$

$\frac{1}{\cos^2x}(\arctan(x))' * y' = x'$

$(arctan(x))' = \frac{x'}{\frac{y'}{\cos^2x}}$

Still not there though :(