Question

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

CALCULUS HELP PLEASE!

Answer 1

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

Answer 2

Alright, so you have:
\[h(\theta)=5\theta\frac{1}{cos(\theta)}+\theta tan(\theta)\]
And you want to find \(h'(\theta)\) RIght?

Answer 3

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

Answer 4

@KeithAfasCalcLover yes!!!

Answer 5

plz some one help i need help plz

Answer 6

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

Answer 7

\[h(\theta)=\theta\left(\frac{5}{cos(\theta)}+tan(\theta)\right)\]
And use the product rule:
\[h'(\theta)=\theta\left(\frac{5}{cos(\theta)}+tan(\theta)\right)'+\left(\frac{5}{cos(\theta)}+tan(\theta)\right)\]
\[h'(\theta)=\theta\left(5\left(\frac{1}{cos(\theta)}\right)'+\frac{1}{cos^2(\theta)}\right)+\left(\frac{5}{cos(\theta)}+tan(\theta)\right)\]

Answer 8

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

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

LOL Well i-i-im flattered hahaha

Answer 17

What Calculus do you have? High School? University?

Answer 18

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

Answer 19

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

Answer 20

@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

Answer 21

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

Answer 22

@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

Answer 23

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

Answer 24

Probably by now, covered limits, and derivatives right?

Answer 25

@KeithAfasCalcLover yea!! i think we are on derivative

Answer 26

@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?:'(

Answer 27

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

Answer 28

@KeithAfasCalcLover yes!!

Answer 29

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)?

Answer 30

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

Answer 31

;(

Answer 32

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

Answer 33

Little bit better?

Answer 34

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

Answer 35

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

Answer 36

Yeah!
So now lets take the derivative of \(g(x)\) and \(h(x)\). Ill do the first and you do the second deal? :-)

Answer 37

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

Answer 38

@KeithAfasCalcLover yay!!! ok deal!;)

Answer 39

Haha done! Your turn

Answer 40

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

Answer 41

@KeithAfasCalcLover did i get it???

Answer 42

NICE haha

Answer 43

@KeithAfasCalcLover wootwooot:D

Answer 44

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!:
\[f'(x)=\left(\frac{x^2}{x^2+1}\right)'=\frac{2x(x^2+1)-2x(x^2)}{(x^2+1)^2}\]

Answer 45

AAAAAND We can simplify a little:
\[f'(x)=\frac{2x^3+2x-2x^3}{(x^2+1)^2}\]
\[f'(x)=\frac{2x}{(x^2+1)^2}\]

Answer 46

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

Answer 47

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? :-)

Answer 48

when x=0????

Answer 49

YEAH. Which makes sense! Because:

Answer 50

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

Answer 51

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

Answer 52

Haha great! Anytime!

Answer 53

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

Answer 54

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

Answer 55

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...