3rd and 4th Derivatives of Trig Fuctions

Hi, can anyone help me with the third and forth derivatives of y=tan x

I get

f(x)=tan x
f '(x)=sec^2 x
f ''(x)=2Sin x . sec^3 x

but am strugling with

f '''(x)=
and
f ''''(x)=

Thanks :smile:

Reply 1

Swayum

Use the product rule.

*Edit*

So you'd have u = 2sinx and v = sec^3x.

Reply 2

bruceleej

Let me guess, you are trying to find the series expansion for tan x.

Reply 3

Saturday Night Special

Correct

Reply 4

nota bene

Yuk, what a question... well use the product and chain rule(I'm ignoring the constant):

f'''(x)=cos(x).sec^3(x)+sin(x).(d/dx)(sec^3(x))

What do you get with the chain rule on differentiating sec^3(x) ?

Reply 5

bruceleej

Yea, you are gonna have to find higher derivatives than that because most of them turn out to be zero when you sub in x=0. Good Luck...

Reply 6

Kolya

I think it is straightforward if you stick to secants and tangents:

f(x) = tanx
f'(x) = sec^2x
f''(x) = 2tanxsec^2x
f'''(x) = 2sec^4x + 4tan^2xsec^2x
f''''(x) = 8tanxsec^4x + 8tanxsec^4x + 8tan^3xsec^2x = 16tanxsec^4x + 8tan^3xsec^2x
etc...

I would not think you need to find higher than the fifth derivative.

Reply 7

generalebriety

Why bother with all this?

We know that sin x = x - x^3/6 + x^5/120 - ..., cos x = 1 - x^2/2 + x^4/24 - ...

So now write tan x like this:

$\tan x = \frac{\sin x}{\cos x} = \frac{x - x^3/6 + x^5/120 + \dots}{1 - x^2/2 + x^4/24 + \dots}$

$\approx (x - x^3/6 + x^5/120)(1 + (- x^2/2 + x^4/24))^{-1}$

and expand the second term using the binomial theorem and multiply out. Also not fun, but a lot more 'mechanical', I find - and therefore less tedious.

Reply 8

nota bene

...or we can say that tan(x) is an odd function and thus we can compare coefficients for

$(Ax+Bx^3+Cx^5+...)(1-\frac{x^2}{2}+\frac{x^4}{4!}-...)=(x-\frac{x^3}{3!}+\frac{x^5}{5!}-...)$
LHS= $Ax+\frac{2B-A}{2}x^3+\frac{24C-12B+A}{24}x^5$
Collect relevant terms: A=1, B=1/3, C=2/15

As you see, there are several methods to do this, which are normally nicer than differentiating.