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Derivative of the sum is the sum of the derivatives.

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

sec²x = tan²x + 1

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

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

= 0 at the end

the answer in the book is zero...

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

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

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

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

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

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

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

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

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

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

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

y=tan(xsinx)

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

sin?

Ahh, but sin itself is inside tangent, no?

sec2x du/dx

ok then wats nxt?

-cosx

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

(xdsinx)[sinxdx]

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

sin2x

yes im following

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

x(-cosx)+sinx(1)

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

-cos x

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

Yeah, derivative of sine is cosine
Proof on request.

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

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

Y=sec4x-tan4x

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

not exactly it is just an answer in the book.

another question sir kirbs
Y=sec4x-tan4x

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

yes

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

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

I hope it's clear

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

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

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

ahh

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

THANK YOU A LOT KIRB...

Your welcome :)

TY KIRB

:)