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

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

Mathematics
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Ooo trig! :O Fun stuff!
Do you know derivative of sec(x)? :)
yes, tan(x)sec(x)

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\[\large\rm \frac{d}{dx}\sec(x)=\sec(x)\tan(x)\]\[\large\rm \frac{d}{dx}\sec(8x)=\sec(8x)\tan(8x)\cdot\color{royalblue}{(8x)'}\]Good good good. So to finish up the derivative of the second term, you need only make sure you apply your chain rule.
so the derivative of sec(8x) is sec(8x)tan(8x)?
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`.
Derivative of 8x? :o
8
\[\large\rm \frac{d}{dx}\sec(8x)=\sec(8x)\tan(8x)\cdot\color{royalblue}{(8x)'}=8\sec(8x)\tan(8x)\] Ok good! :) That takes care of the second term.
An extra 8 popping out due to our chain rule.
chain rule, whenever stuff are inside think of performing that
Remember your cosine derivative? :)
-sin(x)
i mean stuff except that not of the form x only
\[\large\rm \frac{d}{dx}\cos(x)=-\sin(x)\]Good. We'll follow the same pattern, applying chain rule again though.\[\large\rm \frac{d}{dx}\cos\left(\frac{1}{x}\right)=-\sin\left(\frac{1}{x}\right)\cdot\color{royalblue}{\left(\frac{1}{x}\right)'}\]
Do you remember how to differentiate something when it's in the denominator like that? :)
(1/x)' = (x^-1)' = (-x^-2) = (-1/x^2)
correct?
yes good!
\[\large\rm \frac{d}{dx}\cos\left(\frac{1}{x}\right)=-\sin\left(\frac{1}{x}\right)\cdot\color{orangered}{\left(-\frac{1}{x^2}\right)}\]Yayyy good job \c:/ Maybe simplify things down a tad, bring the negatives together and such.\[\large\rm \frac{d}{dx}\cos\left(\frac{1}{x}\right)=\frac{1}{x^2}\cdot\sin\left(\frac{1}{x}\right)\]
Does that need to be simplified any further?
No, not unless you wanted to get a common denominator between the two terms. That seems unnecessary though.
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: \[\frac{d}{dx}cos^{-1}(x)=-1*cos^{-2}(x)*(-sin(x)=tan(x)*cos^{-1}(x)=tan(x)sec(x)\]
so the final answer is \[\frac{ 1 }{ x ^{2} }\sin(\frac{ 1 }{ x })+8\sec(8x)\tan(8x)\] ?
you can do the same with cosec(x), tan(x), etc
yay good job \c:/
oh no there a down side on writing 1/cos that way you will get mixed up with inverse function
just leave it as it is!
or use parenthesis (cosx)^-1
Oh yeah sorry.... implied by the context but your right poor form for those not as comfortable im sorry :/
I usually use arc... but I get your point
yeah but now everyone use arc lol we use them interchangeably
not* everyone
check this derivative carla f(x)=tan(cosx)
see if you mastered the rule lol

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