Question

find f'(x) of f(x)=cos(1/x)+sec(8x)

Answer 1

Ooo trig! :O Fun stuff!

Answer 2

Do you know derivative of sec(x)? :)

Answer 3

yes, tan(x)sec(x)

Answer 4

\[\large\rm \frac{d}{dx}\sec(x)=\sec(x)\tan(x)\]\[\large\rm \frac{d}{dx}\sec(8x)=\sec(8x)\tan(8x)\cdot\color{royalblue}{(8x)'}\]Good good good.
So to finish up the derivative of the second term,
you need only make sure you apply your chain rule.

Answer 5

so the derivative of sec(8x) is sec(8x)tan(8x)?

Answer 6

Not entirely :O
That's most of it.
Do you see the blue portion though?
You have to multiply by the derivative of the `inner function`.

Answer 7

Derivative of 8x? :o

Answer 8

8

Answer 9

\[\large\rm \frac{d}{dx}\sec(8x)=\sec(8x)\tan(8x)\cdot\color{royalblue}{(8x)'}=8\sec(8x)\tan(8x)\]
Ok good! :) That takes care of the second term.

Answer 10

An extra 8 popping out due to our chain rule.

Answer 11

chain rule, whenever stuff are inside think of performing that

Answer 12

Remember your cosine derivative? :)

Answer 13

-sin(x)

Answer 14

i mean stuff except that not of the form x only

Answer 15

\[\large\rm \frac{d}{dx}\cos(x)=-\sin(x)\]Good.
We'll follow the same pattern, applying chain rule again though.\[\large\rm \frac{d}{dx}\cos\left(\frac{1}{x}\right)=-\sin\left(\frac{1}{x}\right)\cdot\color{royalblue}{\left(\frac{1}{x}\right)'}\]

Answer 16

Do you remember how to differentiate something when it's in the denominator like that? :)

Answer 17

(1/x)' = (x^-1)' = (-x^-2) = (-1/x^2)

Answer 18

correct?

Answer 19

yes good!

Answer 20

\[\large\rm \frac{d}{dx}\cos\left(\frac{1}{x}\right)=-\sin\left(\frac{1}{x}\right)\cdot\color{orangered}{\left(-\frac{1}{x^2}\right)}\]Yayyy good job \c:/
Maybe simplify things down a tad, bring the negatives together and such.\[\large\rm \frac{d}{dx}\cos\left(\frac{1}{x}\right)=\frac{1}{x^2}\cdot\sin\left(\frac{1}{x}\right)\]

Answer 21

Does that need to be simplified any further?

Answer 22

No, not unless you wanted to get a common denominator between the two terms.
That seems unnecessary though.

Answer 23

Just thought I would add the trick to remember the derivative to those nastier trig functions: sec(x)=1/cos(x):
\[cos^{-1}(x)\]
and therefore by the chain rule:
\[\frac{d}{dx}cos^{-1}(x)=-1*cos^{-2}(x)*(-sin(x)=tan(x)*cos^{-1}(x)=tan(x)sec(x)\]

Answer 24

so the final answer is \[\frac{ 1 }{ x ^{2} }\sin(\frac{ 1 }{ x })+8\sec(8x)\tan(8x)\] ?

Answer 25

you can do the same with cosec(x), tan(x), etc

Answer 26

yay good job \c:/

Answer 27

oh no there a down side on writing 1/cos that way
you will get mixed up with inverse function

Answer 28

just leave it as it is!

Answer 29

or use parenthesis (cosx)^-1

Answer 30

Oh yeah sorry.... implied by the context but your right poor form for those not as comfortable im sorry :/

Answer 31

I usually use arc... but I get your point

Answer 32

yeah but now everyone use arc lol
we use them interchangeably

Answer 33

not* everyone

Answer 34

check this derivative carla f(x)=tan(cosx)

Answer 35

see if you mastered the rule lol

Answer 36

@clara1223