find the derivative of the trigonometric function…
h(theta)=5 theta sec theta+theta tan theta
CALCULUS HELP PLEASE!

- anonymous

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- anonymous

please help me!!!!!!!!!!!

- 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

keith plz help man on my question a fewer more and than i got to go tot bed

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## More answers

- anonymous

@KeithAfasCalcLover yes!!!

- Candy13106

plz some one help i need help plz

- anonymous

group theta like this:\[h(\theta) = \theta*(5*\sec(\theta)+\tan(\theta)\]

- 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

\[h'(\theta)=\theta\left(5\left(\frac{-1}{cos^2(\theta)}\times-sin(\theta)\right)+\frac{1}{cos^2(\theta)}\right)+\left(\frac{5}{cos(\theta)}+tan(\theta)\right)\]
\[h'(\theta)=\theta\left(\frac{5sin(\theta)+1}{cos^2(\theta)}\right)+\left(\frac{5}{cos(\theta)}+tan(\theta)\right)\]

- anonymous

\[h'(\theta)=\frac{\theta[5sin(\theta)+1]}{cos^2(\theta)}+\frac{5+sin(\theta)}{cos(\theta)}\]

- anonymous

I probably could go a lil bit further but ill stop there. Cool Hannah?

- anonymous

@KeithAfasCalcLover omggg thank you so muchh!!!!!! you are the best

- anonymous

Haha I try ;-) Im glad you liked it. Anytime.

- anonymous

@KeithAfasCalcLover couldi ask you some more questions later on? ive been having so much trouble doing this stupid hw:'(

- anonymous

Yeah sure! Whenever i'm on i'd love to answer some Calc questions! HAVE YOU SEEN MY NAME? Haha

- anonymous

@KeithAfasCalcLover hahaha omg totally didnt notice that until now;) i think i really like that name <333

- anonymous

LOL Well i-i-im flattered hahaha

- anonymous

What Calculus do you have? High School? University?

- anonymous

@KeithAfasCalcLover haha;) im taking calculus ab online so i guess it's highschool calc:)

- anonymous

Ahh I see. Just out of curiosity, why online? If you don't mind me asking :p

- 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

Haha naww don't worry. Ehh whats bc calculus and ab calculus? Im not familiar with the terminology haha

- 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

Lol ahh I see. So how far did you guys get so far?

- anonymous

Probably by now, covered limits, and derivatives right?

- anonymous

@KeithAfasCalcLover yea!! i think we are on derivative

- 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

haha its not that bad. The "horizontal line" means that the line has a slope of zero right?

- anonymous

@KeithAfasCalcLover yes!!

- 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

@KeithAfasCalcLover g(x) is x?? im not suree

- anonymous

;(

- anonymous

Lol aww don't cry...
\[g(x)=numerator\]
\[h(x)=denominator\]

- anonymous

Little bit better?

- anonymous

So therefore \(g(x)=x^2\) and \(h(x)\) would be?

- anonymous

@KeithAfasCalcLover will it be x^2+1??

- 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

\[g'(x)=(x^2)'=2x\]

- anonymous

@KeithAfasCalcLover yay!!! ok deal!;)

- anonymous

Haha done! Your turn

- anonymous

h'(x)=2x..........?

- anonymous

@KeithAfasCalcLover did i get it???

- anonymous

NICE haha

- anonymous

@KeithAfasCalcLover wootwooot:D

- 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

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

But we're still not done. We need to find out when \(f'(x)=0\)

- 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

when x=0????

- anonymous

YEAH. Which makes sense! Because:

##### 1 Attachment

- anonymous

The tangent line at \(x=0\) would be horizontal, you can see!

- anonymous

@KeithAfasCalcLover omggg!! yes i get it!! THANKS SO MUCHH

- anonymous

Haha great! Anytime!

- anonymous

Yeahhh. Im in grade 12 now! Ill be getting the first introduction of calculus in like March haha

- anonymous

@KeithAfasCalcLover omg you go to highschool? how are you soo good at math?!?

- 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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