anonymous
  • anonymous
find the derivative of the trigonometric function… h(theta)=5 theta sec theta+theta tan theta CALCULUS HELP PLEASE!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
please help me!!!!!!!!!!!
anonymous
  • anonymous
Alright, so you have: \[h(\theta)=5\theta\frac{1}{cos(\theta)}+\theta tan(\theta)\] And you want to find \(h'(\theta)\) RIght?
Candy13106
  • Candy13106
keith plz help man on my question a fewer more and than i got to go tot bed

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anonymous
  • anonymous
@KeithAfasCalcLover yes!!!
Candy13106
  • Candy13106
plz some one help i need help plz
anonymous
  • anonymous
group theta like this:\[h(\theta) = \theta*(5*\sec(\theta)+\tan(\theta)\]
anonymous
  • anonymous
\[h(\theta)=\theta\left(\frac{5}{cos(\theta)}+tan(\theta)\right)\] And use the product rule: \[h'(\theta)=\theta\left(\frac{5}{cos(\theta)}+tan(\theta)\right)'+\left(\frac{5}{cos(\theta)}+tan(\theta)\right)\] \[h'(\theta)=\theta\left(5\left(\frac{1}{cos(\theta)}\right)'+\frac{1}{cos^2(\theta)}\right)+\left(\frac{5}{cos(\theta)}+tan(\theta)\right)\]
anonymous
  • anonymous
\[h'(\theta)=\theta\left(5\left(\frac{-1}{cos^2(\theta)}\times-sin(\theta)\right)+\frac{1}{cos^2(\theta)}\right)+\left(\frac{5}{cos(\theta)}+tan(\theta)\right)\] \[h'(\theta)=\theta\left(\frac{5sin(\theta)+1}{cos^2(\theta)}\right)+\left(\frac{5}{cos(\theta)}+tan(\theta)\right)\]
anonymous
  • anonymous
\[h'(\theta)=\frac{\theta[5sin(\theta)+1]}{cos^2(\theta)}+\frac{5+sin(\theta)}{cos(\theta)}\]
anonymous
  • anonymous
I probably could go a lil bit further but ill stop there. Cool Hannah?
anonymous
  • anonymous
@KeithAfasCalcLover omggg thank you so muchh!!!!!! you are the best
anonymous
  • anonymous
Haha I try ;-) Im glad you liked it. Anytime.
anonymous
  • anonymous
@KeithAfasCalcLover couldi ask you some more questions later on? ive been having so much trouble doing this stupid hw:'(
anonymous
  • anonymous
Yeah sure! Whenever i'm on i'd love to answer some Calc questions! HAVE YOU SEEN MY NAME? Haha
anonymous
  • anonymous
@KeithAfasCalcLover hahaha omg totally didnt notice that until now;) i think i really like that name <333
anonymous
  • anonymous
LOL Well i-i-im flattered hahaha
anonymous
  • anonymous
What Calculus do you have? High School? University?
anonymous
  • anonymous
@KeithAfasCalcLover haha;) im taking calculus ab online so i guess it's highschool calc:)
anonymous
  • anonymous
Ahh I see. Just out of curiosity, why online? If you don't mind me asking :p
anonymous
  • anonymous
@KeithAfasCalcLover totally dont mind! im taking bc at school and i thought taking it ab online will help me to review stuff twice and boost up my gpa...but now i kinda regret taking it with all these overwhelming problems:O
anonymous
  • anonymous
Haha naww don't worry. Ehh whats bc calculus and ab calculus? Im not familiar with the terminology haha
anonymous
  • anonymous
@KeithAfasCalcLover they are just highschool calc and bc is a little harder than ab and covers more topics but they are basically the same thing:p
anonymous
  • anonymous
Lol ahh I see. So how far did you guys get so far?
anonymous
  • anonymous
Probably by now, covered limits, and derivatives right?
anonymous
  • anonymous
@KeithAfasCalcLover yea!! i think we are on derivative
anonymous
  • anonymous
@KeithAfasCalcLover determine the point(s) at which the graph of the function has a horizontal tangent line. f(x)=x^2/(x^2+1) how do you do this one?:'(
anonymous
  • anonymous
haha its not that bad. The "horizontal line" means that the line has a slope of zero right?
anonymous
  • anonymous
@KeithAfasCalcLover yes!!
anonymous
  • anonymous
So then the "horizontal tangent line" can be found at the x-values where the derivative is zero since the derivative describes the instantaneous slopes of the function at any point! So you have : \[f(x)=\frac{x^2}{x^2+1}\] You can find the derivative by applying the quotient rule which says: \[f'(x)=\left(\frac{g(x)}{h(x)}\right)'=\frac{g'(x)h(x)-h'(x)g(x)}{[h(x)]^2}\] So lets break it down: What is our g(x) and h(x)?
anonymous
  • anonymous
@KeithAfasCalcLover g(x) is x?? im not suree
anonymous
  • anonymous
;(
anonymous
  • anonymous
Lol aww don't cry... \[g(x)=numerator\] \[h(x)=denominator\]
anonymous
  • anonymous
Little bit better?
anonymous
  • anonymous
So therefore \(g(x)=x^2\) and \(h(x)\) would be?
anonymous
  • anonymous
@KeithAfasCalcLover will it be x^2+1??
anonymous
  • anonymous
Yeah! So now lets take the derivative of \(g(x)\) and \(h(x)\). Ill do the first and you do the second deal? :-)
anonymous
  • anonymous
\[g'(x)=(x^2)'=2x\]
anonymous
  • anonymous
@KeithAfasCalcLover yay!!! ok deal!;)
anonymous
  • anonymous
Haha done! Your turn
anonymous
  • anonymous
h'(x)=2x..........?
anonymous
  • anonymous
@KeithAfasCalcLover did i get it???
anonymous
  • anonymous
NICE haha
anonymous
  • anonymous
@KeithAfasCalcLover wootwooot:D
anonymous
  • anonymous
So we know: \(g(x)=x^2\) \(g'(x)=2x\) \(h(x)=x^2+1\) \(h'(x)=2x\) And we know that: \[f'(x)=\left(\frac{g(x)}{h(x)}\right)'=\frac{g'(x)h(x)-h'(x)g(x)}{[h(x)]^2}\] So lets plug everything in!: \[f'(x)=\left(\frac{x^2}{x^2+1}\right)'=\frac{2x(x^2+1)-2x(x^2)}{(x^2+1)^2}\]
anonymous
  • anonymous
AAAAAND We can simplify a little: \[f'(x)=\frac{2x^3+2x-2x^3}{(x^2+1)^2}\] \[f'(x)=\frac{2x}{(x^2+1)^2}\]
anonymous
  • anonymous
But we're still not done. We need to find out when \(f'(x)=0\)
anonymous
  • anonymous
So lets set it to zero: \[0=\frac{2x}{(x^2+1)^2}\] \[0(x^2+1)^2=2x\] \[0=2x\] \[0/2=x\] \[x=0\] So when will there be a horizontal tangent line? :-)
anonymous
  • anonymous
when x=0????
anonymous
  • anonymous
YEAH. Which makes sense! Because:
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anonymous
  • anonymous
The tangent line at \(x=0\) would be horizontal, you can see!
anonymous
  • anonymous
@KeithAfasCalcLover omggg!! yes i get it!! THANKS SO MUCHH
anonymous
  • anonymous
Haha great! Anytime!
anonymous
  • anonymous
Yeahhh. Im in grade 12 now! Ill be getting the first introduction of calculus in like March haha
anonymous
  • anonymous
@KeithAfasCalcLover omg you go to highschool? how are you soo good at math?!?
anonymous
  • anonymous
Lol haha Well I have a problem. If I have a question about something or something I need to know, I find it out haha. So I heard of calculus in grade 11 and heard it's applications. So I got curious and looked at derivatives and limits. Then I heard of the idea of area under a curve and found that it was related to the integral so I learned that. Then I learned ways to apply it to multivariable approaches so like multivariable calculus. And that's where im at right now. I always had a thing for math. It comes to me like English comes to people who speak. Its really like a language haha. yeahhhh...

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