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clara1223
 one year ago
find f'(x) of f(x)=cos(1/x)+sec(8x)
clara1223
 one year ago
find f'(x) of f(x)=cos(1/x)+sec(8x)

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zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Ooo trig! :O Fun stuff!

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Do you know derivative of sec(x)? :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3\[\large\rm \frac{d}{dx}\sec(x)=\sec(x)\tan(x)\]\[\large\rm \frac{d}{dx}\sec(8x)=\sec(8x)\tan(8x)\cdot\color{royalblue}{(8x)'}\]Good good good. So to finish up the derivative of the second term, you need only make sure you apply your chain rule.

clara1223
 one year ago
Best ResponseYou've already chosen the best response.0so the derivative of sec(8x) is sec(8x)tan(8x)?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Not 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`.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3\[\large\rm \frac{d}{dx}\sec(8x)=\sec(8x)\tan(8x)\cdot\color{royalblue}{(8x)'}=8\sec(8x)\tan(8x)\] Ok good! :) That takes care of the second term.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3An extra 8 popping out due to our chain rule.

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0chain rule, whenever stuff are inside think of performing that

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Remember your cosine derivative? :)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0i mean stuff except that not of the form x only

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3\[\large\rm \frac{d}{dx}\cos(x)=\sin(x)\]Good. We'll follow the same pattern, applying chain rule again though.\[\large\rm \frac{d}{dx}\cos\left(\frac{1}{x}\right)=\sin\left(\frac{1}{x}\right)\cdot\color{royalblue}{\left(\frac{1}{x}\right)'}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Do you remember how to differentiate something when it's in the denominator like that? :)

clara1223
 one year ago
Best ResponseYou've already chosen the best response.0(1/x)' = (x^1)' = (x^2) = (1/x^2)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3\[\large\rm \frac{d}{dx}\cos\left(\frac{1}{x}\right)=\sin\left(\frac{1}{x}\right)\cdot\color{orangered}{\left(\frac{1}{x^2}\right)}\]Yayyy good job \c:/ Maybe simplify things down a tad, bring the negatives together and such.\[\large\rm \frac{d}{dx}\cos\left(\frac{1}{x}\right)=\frac{1}{x^2}\cdot\sin\left(\frac{1}{x}\right)\]

clara1223
 one year ago
Best ResponseYou've already chosen the best response.0Does that need to be simplified any further?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3No, not unless you wanted to get a common denominator between the two terms. That seems unnecessary though.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Just 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: \[\frac{d}{dx}cos^{1}(x)=1*cos^{2}(x)*(sin(x)=tan(x)*cos^{1}(x)=tan(x)sec(x)\]

clara1223
 one year ago
Best ResponseYou've already chosen the best response.0so the final answer is \[\frac{ 1 }{ x ^{2} }\sin(\frac{ 1 }{ x })+8\sec(8x)\tan(8x)\] ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you can do the same with cosec(x), tan(x), etc

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0oh no there a down side on writing 1/cos that way you will get mixed up with inverse function

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0just leave it as it is!

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0or use parenthesis (cosx)^1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh yeah sorry.... implied by the context but your right poor form for those not as comfortable im sorry :/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I usually use arc... but I get your point

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0yeah but now everyone use arc lol we use them interchangeably

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0not* everyone

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0check this derivative carla f(x)=tan(cosx)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0see if you mastered the rule lol

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0@clara1223
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