how about differentiate: y=sec²x-tan²x

- anonymous

how about differentiate: y=sec²x-tan²x

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- schrodinger

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- terenzreignz

Derivative of the sum is the sum of the derivatives.

- terenzreignz

Although, to be really creative, you can just recall that the derivative of a constant is 0
:D

- terenzreignz

sec²x = tan²x + 1

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## More answers

- anonymous

\[
\sec^2x-\tan^2x = \frac{1}{\cos^2x} - \frac{\sin^2 x}{\cos^2 x}
\]

- anonymous

So I mean you're gonna end up with y=1

- kirbykirby

\[\frac{d}{dx}(\sec^2x-\tan^2x)=2\sec x(\sec x \tan x)-2\tan x(\sec^2x) = 2\sec^2x tanx - 2\sec^2x \tan x = 0 \]

- kirbykirby

= 0 at the end

- anonymous

the answer in the book is zero...

- tkhunny

Please listen to @terenzreignz
y = 1 Don't miss these sipmle ones.

- kirbykirby

Sorry it cut off a but, but the last one just cut off "tan x"

- terenzreignz

So many different ways to do it, but if all else eludes you, do it directly, as @kirbykirby
But
sec²x = tan²x + 1
So
sec²x - tan²x = tan²x + 1 - tan²x = 1
And the derivative of 1...?

- kirbykirby

Yes there are many ways to approach this one. You can re-write sec^x and tan^x in terms of cos x and sin x, but if you know the derivative of sec x and tan x (I recommend that you t do), then it's easier to proceed that way

- anonymous

i will get the double prime if i will get the derivative 1...?

- terenzreignz

It might pay (and save you time) to know by heart your trigonometric identities, too.

- tkhunny

It absolutely WILL pay huge dividends to brush up on your trigonometry. If you were taking my examine and you did ANYTHING on this problem except recognize IMMEDIATELY that \(\dfrac{d}{dt}[\sec^{2}(t) - \tan^{2}(t)\ = \dfrac{d}{dt}1 = 0\), I would consider recommending you for a refresher course in a lower level class. Gotta know your trig. It's not optional!

- kirbykirby

@jayshane I'm not sure what you mean about getting a double prime? But the derivative of 1 (or any constant) is just 0 :)

- kirbykirby

Ok don't get so uptight @tkhunny :S Oye. It's not THE end of the world to not see it immediately o_o. I;m past cal 3 and doing differential equations now and even I didn't see it right away as you mentioned it lool. And I mean I barely use trig anymore that I am doing stats now

- kirbykirby

I usually only think of these identities when I can't "plug and chug" right away lol

- tkhunny

@kirbykirby No uptight going on here, but I do appreciate your calming influence. I feel very strongly about this particular issue. One should not be in differential equations without a very solid foundation in trigonometry. Simple as that.
Also, as we have yet to see ANY effort by @jayshane, it's just not clear what we're doing, here. A little strong reality therapy seemed like it might be appropriate.

- kirbykirby

Well I don't think that's a fair assumption to make that I don't have a good foundation in trig because I didn't recognize that as THE first thing to do?! o_o

- anonymous

i acctually having a hard time dealing with trigonometric identities..
and im studying harder and harder on it its just im just an ordinary man..
but im really tring hard pls dont be anger on me...

- terenzreignz

Nobody's angry with you @jayshane
But, if you have the time, do study trigonometry... :)

- kirbykirby

No we are not angry at you :) We were just recommending to really know the trig identities as much as you can because they save you time :)

- anonymous

ive been memorizing it but i dont know how to use them in solving..

- tkhunny

@ jayshane No anger, just encouragement. Get a trig book and work all 4,000 problems. Get it on your head and it will pay great dividends. @kirbykirby seems to be a good case that it isn't absolutely necessary, but if you could spend ALL your time in Differential Equations studying Differential Equations, don't you think this would be better than spending 1/2 your time in Differential Equations trying to remember trigonometry?

- anonymous

ok thank you for the answers i will study more...

- anonymous

but for now please help me just last 2 questions..!!

- tkhunny

Good call. Show your work, too. This will help us as much as it helps you.

- kirbykirby

Well I am not spending 1/2 my time remembering trig :P I do remember the identities but I just don't necessarily apply them right away unless I am stuck, or just force myself to be "oh maybe I should use a trig identity here". Do you know what I mean?

- anonymous

my teacher encourages us to use the long method and i just cant get it,,

- kirbykirby

And anyway even if you do forget, it's very easy to pull up a cal book and look for the identities, (although you should know them for the exam!!)

- tkhunny

@kirbykirby Right. No ridicule intended. "1/2" was a bit hyperbolic. There are many survival methodologies. I was kind of a freak in my first differential equations class, so that does tend to make me think my survival method is the only way to go. I do realize there are humans on this planet. :-)

- terenzreignz

@jayshane
Your teacher recommends the long method for your practice
It's not on every opportunity that knowledge of trigonometric identities will save you from long calculations, BUT
the long method is the only method that would work for all situations...
Now, you said you have two more questions?

- kirbykirby

@tkhunny Ok good to know we are on good terms :P As for @jayshane, you still need help right

- anonymous

y=tan(xsinx)

- tkhunny

@jayshane I can think of only one reason to encourage a class to use the "long method". Your teacher knows your differential calculus isn't very good and is forcing you to practice! :-)

- terenzreignz

Chain rule, @jayshane
A good way to work on this is to identify what you could call the "outermost" function, the function that sort of "contains" all the others. Which function seems to fit this description?

- anonymous

she does just give this assignment to us and she doesnt even give explanations she just gave the identities and were having a hard time to deal with it..!

- terenzreignz

Well, @jayshane
That's why we're here :)
Now, which function is outermost?

- anonymous

sin?

- terenzreignz

Ahh, but sin itself is inside tangent, no?

- kirbykirby

Yeah A good trick for the chain rule as @terenzreignz mentions, identity your outer functions and inner functions for the chain rule. Think of what the elementary function is and what changed:
Here, we could think of tan x being the elementary function, but we changed x to "xsin x". So, you can think of it like a box: you have xsinx in the box tan[ ] , so you find the derivative of tan [ ] and multiply by the derivative inside the box [ ]

- terenzreignz

y=tan(xsinx)
And I've already named the outermost function...
So you know tan contains all others, what's the derivative of tan?

- anonymous

sec2x du/dx

- terenzreignz

leave out the du/dx, and stick to
sec² x
So "write that down"
except instead of x, copy whatever was "inside" the tan function, namely, xsin x
(NOTE: We're not yet done!)

- tkhunny

\(f(x) = \tan(x\cdot\sin(x))\) -- What a fascinating function. I'm not sure I've ever seen that one in a problem set.

- anonymous

ok then wats nxt?

- anonymous

Remember the Pythagorean identity? \(\cos^2x+\sin^2x=1\); if you divide through by \(\cos^2x\), it yields \(1+\tan^2x=\sec^2x\). We can apply this identity to simplify our function!
$$y=\sec^2x-\tan^2x=\tan^2x+1-\tan^2x=1$$Do you know what the derivative of a constant function is? What is the slope of a horizontal line?

- terenzreignz

Well, I like to think of chain rule as sort of a hitman's list, and think of it this way, you've already "taken 'tan' out of the picture" so you go for its "inner circle"
lol
anyway
So now that you've "dealt with" tan (lol)
time to deal with whatever was inside it.
So remember
sec²(xsin x)
?
You multiply this to whatever the derivative of the inner function is, namely, xsin x.
So what's the derivative of xsin x?

- kirbykirby

@oldrin.bataku We have covered the answer for that already :) We are approaching a new question. Sorry to burst your bubble

- anonymous

-cosx

- terenzreignz

No... remember
xsin x
is the product of two functions, now would be a great time to recall your product rule?

- kirbykirby

That\s not exactly right. You have x*sin x, not just "sin x" In fact you have a product rule with functions "x" and "sin x" being multiplied

- kirbykirby

d/dx (x*sinx) = (1)sinx + x(cosx) by the product rule

- anonymous

@kirbykirby I posted to explain the identity because none of you had.
@jayshane Why don't we break our function into pieces? and *chain* the derivatives together? $$u=x\sin x\\y=\tan v\\\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=(-\sec^2u)(\sin x+x\cos x)=-\sec^2(x\sin x)[\sin x+x \cos x]$$

- anonymous

(xdsinx)[sinxdx]

- terenzreignz

@oldrin.bataku
I think I did post the identity, though I didn't derive it.
@jayshane
I can see where you're getting at, for as long as
dsin x is
the derivative of sin x
and dx
is the derivative of x.
What's the derivative of sin x?
What's the derivative of x?

- terenzreignz

Also, you don't multiply them, you add them.

- anonymous

sin2x

- terenzreignz

Hang on, let me rewrite your statement properly, in a way that not even your teacher can find an error in :)
\[x\frac{d}{dx}\sin x +(\sin x)\frac{d}{dx}x\]
Now work out these derivatives.

- anonymous

@jayshane the product rule tells us \(\frac{d}{dx}uv=\frac{du}{dx}v+\frac{dv}{dx}u\) -- see how we multiply one of the original functions by the derivative of the other?.

- anonymous

yes im following

- anonymous

In our case of \(\frac{d}{dx}x\sin x\), we have two functions being multiplied: \(x\) and \(\sin x\). Do you know how to determine the derivative of this product using the product rule?

- kirbykirby

I actually find it easier to forget the "u" notation when first learning the chain rule. Just look at it as outer and inner functions. And you won't ever need the u-notation once you master it that way

- terenzreignz

@kirbykirby
This is product rule now, though...
Eventually we do do away with the u and v, probably

- anonymous

@oldrin.bataku yeah

- kirbykirby

Oops I meant the product rule LOL sorry. I am responding to another post and I am getting confused between both posts

- anonymous

x(-cosx)+sinx(1)

- terenzreignz

Almost...
What's the derivative of sin x?

- anonymous

-cos x

- kirbykirby

No not quite. d/dx (sin x) = cos x
d/dx (cos x ) = -sin x... yes it's confusing at first

- terenzreignz

Yeah, derivative of sine is cosine
Proof on request.

- anonymous

can u give me appropriate solution guys, then ill just study it..

- terenzreignz

Well, if you accept that the derivative of sin x is cos x, then
you can work out (PLEASE DO WORK IT OUT)
That the derivative of
xsin x
is
x(cos x) + sin x
Which as a whole expression, you multiply that one we got earlier, which was
sec²(xsin x)
Hence, you get
[sec²(xsin x)][x(cos x) + sin x]

- anonymous

then how will i continue to solve? the answer in the book is 4tanxsec2x

- anonymous

Y=sec4x-tan4x

- kirbykirby

Does your prof require that form of an answer (4tanx sec2x)?

- anonymous

not exactly it is just an answer in the book.

- kirbykirby

Because I can guarantee you it's the answer :) http://www.wolframalpha.com/input/?i=d%2Fdx+tan%28xsinx%29

- anonymous

another question sir kirbs
Y=sec4x-tan4x

- kirbykirby

I tried equation your textbook answer with that one and I don't get an equality... which means the book might have a typo. Either way, it is way too much work to get to an answer like that :)

- kirbykirby

You can proceed in a similar way to what I did. Or, you can make use of the identity:
1+tan^2 x = sec^2 x
So:

- kirbykirby

\[\sec^4x-\tan^4x = (\sec^2x)^2-\tan^4x\]
can you see that?

- anonymous

yes

- kirbykirby

\[=(1+\tan^2x)^2-\tan^4x\]
\[=(1+2\tan^2x+\tan^4x)-\tan^4x\] by using the identity (a+b)^2 = (a+2ab+b^2)

- kirbykirby

\[=1+2\tan^2x\]now our problem is much simpler! We find \[\frac{d}{dx}(1+2\tan^2x)=\frac{d}{dx}1+\frac{d}{dx}2\tan^2x\]

- kirbykirby

\[=0+2\frac{d}{dx}\tan^2x\]
\[=2*(2*\tan x)(\sec^2 x) \] since this is the famous chain rule again. You put down the ^2 from tan( ) in front, and you multiply by the derivative of tan x which is sec^2 x

- kirbykirby

\[=4\tan x \sec^2x\]

- kirbykirby

Think of your box: you have \[[\tan x]^2 = 2[\tan x]*\frac{d}{dx}[\tan x]\]

- kirbykirby

I hope it's clear

- kirbykirby

In general for a "power" you have like \[[f(x)]^n=n[f(x)]^{n-1}*\frac{d}{dx}f(x)\]
I hope I am not confusing you by writing this

- anonymous

=(1+2tan2x+tan4x)−tan4x how did u get this kirb?

- kirbykirby

There should be a derivative d/dx in front of [f(x)]^n, sorry my computer was lagging

- kirbykirby

Ok you know that \[(a+b)^2 = (a+2ab+b^2)\]

- anonymous

ahh

- kirbykirby

\[(1+\tan^2x)^2=(1+2(1)(\tan^2x)+(\tan^2x)^2) \].... here we have that \[a = 1, b = \tan^2x\]

- anonymous

THANK YOU A LOT KIRB...

- kirbykirby

Your welcome :)

- kirbykirby

Using the identity \[1+\tan^2x=\sec^2x\]We have
\[\sec^2x-\tan^2x=(1+\tan^2x)-\tan^2x\]
\[=1\]
So, \[\frac{d}{dx}1=0\]

- anonymous

TY KIRB

- kirbykirby

:)

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