clara1223
  • clara1223
find f'(x) of f(x)=cos(1/x)+sec(8x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
zepdrix
  • zepdrix
Ooo trig! :O Fun stuff!
zepdrix
  • zepdrix
Do you know derivative of sec(x)? :)
clara1223
  • clara1223
yes, tan(x)sec(x)

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zepdrix
  • zepdrix
\[\large\rm \frac{d}{dx}\sec(x)=\sec(x)\tan(x)\]\[\large\rm \frac{d}{dx}\sec(8x)=\sec(8x)\tan(8x)\cdot\color{royalblue}{(8x)'}\]Good good good. So to finish up the derivative of the second term, you need only make sure you apply your chain rule.
clara1223
  • clara1223
so the derivative of sec(8x) is sec(8x)tan(8x)?
zepdrix
  • zepdrix
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`.
zepdrix
  • zepdrix
Derivative of 8x? :o
clara1223
  • clara1223
8
zepdrix
  • zepdrix
\[\large\rm \frac{d}{dx}\sec(8x)=\sec(8x)\tan(8x)\cdot\color{royalblue}{(8x)'}=8\sec(8x)\tan(8x)\] Ok good! :) That takes care of the second term.
zepdrix
  • zepdrix
An extra 8 popping out due to our chain rule.
xapproachesinfinity
  • xapproachesinfinity
chain rule, whenever stuff are inside think of performing that
zepdrix
  • zepdrix
Remember your cosine derivative? :)
clara1223
  • clara1223
-sin(x)
xapproachesinfinity
  • xapproachesinfinity
i mean stuff except that not of the form x only
zepdrix
  • zepdrix
\[\large\rm \frac{d}{dx}\cos(x)=-\sin(x)\]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
  • zepdrix
Do you remember how to differentiate something when it's in the denominator like that? :)
clara1223
  • clara1223
(1/x)' = (x^-1)' = (-x^-2) = (-1/x^2)
clara1223
  • clara1223
correct?
xapproachesinfinity
  • xapproachesinfinity
yes good!
zepdrix
  • zepdrix
\[\large\rm \frac{d}{dx}\cos\left(\frac{1}{x}\right)=-\sin\left(\frac{1}{x}\right)\cdot\color{orangered}{\left(-\frac{1}{x^2}\right)}\]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
  • clara1223
Does that need to be simplified any further?
zepdrix
  • zepdrix
No, not unless you wanted to get a common denominator between the two terms. That seems unnecessary though.
anonymous
  • anonymous
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: \[\frac{d}{dx}cos^{-1}(x)=-1*cos^{-2}(x)*(-sin(x)=tan(x)*cos^{-1}(x)=tan(x)sec(x)\]
clara1223
  • clara1223
so the final answer is \[\frac{ 1 }{ x ^{2} }\sin(\frac{ 1 }{ x })+8\sec(8x)\tan(8x)\] ?
anonymous
  • anonymous
you can do the same with cosec(x), tan(x), etc
zepdrix
  • zepdrix
yay good job \c:/
xapproachesinfinity
  • xapproachesinfinity
oh no there a down side on writing 1/cos that way you will get mixed up with inverse function
xapproachesinfinity
  • xapproachesinfinity
just leave it as it is!
xapproachesinfinity
  • xapproachesinfinity
or use parenthesis (cosx)^-1
anonymous
  • anonymous
Oh yeah sorry.... implied by the context but your right poor form for those not as comfortable im sorry :/
anonymous
  • anonymous
I usually use arc... but I get your point
xapproachesinfinity
  • xapproachesinfinity
yeah but now everyone use arc lol we use them interchangeably
xapproachesinfinity
  • xapproachesinfinity
not* everyone
xapproachesinfinity
  • xapproachesinfinity
check this derivative carla f(x)=tan(cosx)
xapproachesinfinity
  • xapproachesinfinity
see if you mastered the rule lol
xapproachesinfinity
  • xapproachesinfinity
@clara1223

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