Question

If y=| cos x - sin x |, find f'(90 degree)

Answer 1

@zepdrix

Answer 2

This one is a little annoying.
You can either remember this little fact:\[\Large\bf\sf \frac{d}{dx}|x|\quad=\quad \frac{x}{|x|}\]Or we can work it out the long way if necessary.

Answer 3

please provide me the long way...... I want to learn it from....the longer one

Answer 4

\[\Large\bf\sf |x|\quad:=\sqrt{x^2}\]So we'll start with this definition and take the derivative,

Answer 5

\[\Large\bf\sf \left(\sqrt {x^2}\right)'\quad=\quad\frac{1}{2\sqrt{x^2}}\cdot(x^2)'\]

Answer 6

Understand what I did there? :o
Do you remember the derivative of sqrt(x) ?
It's a good one to have memorized.

Answer 7

yes got it till here, GO ON

Answer 8

So if we cancel the 2's out,\[\Large\bf\sf \left(\sqrt {x^2}\right)'\quad=\quad\frac{x}{\sqrt{x^2}}\]And recall our denominator was defined to be |x| before, so we'll put it back in now.\[\Large\bf\sf |x|'\quad=\quad\frac{x}{|x|}\]

Answer 9

Mmm I guess that wasn't super long XD lol.
With this specific problem you have though, it might be a little weird when we go to evaluate it at 90 degrees. We'll need to be careful.

Answer 10

ok then do it in radian .... I can convert it

Answer 11

1) find f(90 degrees)
f(90 degrees) = |cos(x)-sin(x)|=sin(x)-cos(x)=1
2) take the derivative of sin(x)-cos(x)
3) plug in for x=90 degrees

Answer 12

Explain this step \Large\bf\sf |x|\quad:=\sqrt{x^2}

Answer 13

How to use the theory taught by you here ?

Answer 14

Ummm I like eashy's method a little better.
You have less stuff floating around.

At 90 degrees, our function |cos x-sin x| = -(cos x-sin x) = sin x - cos x.
And it's much easier to take a derivative from there.

The way I was doing it, we would apply the derivative,\[\Large\bf\sf |\cos x-\sin x|'\quad=\quad \frac{\cos x - \sin x}{|\cos x - \sin x|}\cdot \left(\cos x - \sin x\right)'\]We get the derivative of the absolute, but then we have to apply the chain rule.
This should still work, it's just a bit burdensome with all of these pieces.

Answer 15

Please explain this step


|cos x-sin x| = -(cos x-sin x) = sin x - cos x

Answer 16

Here is another way we can define absolute value, as a piece-wise function,\[\Large |x|=\cases{\quad x,\quad x>0\\-x, \quad x<0}\]

So applying this to our problem,\[\Large |\cos x-\sin x|=\cases{\quad \cos x-\sin x,\qquad \cos x-\sin x>0\\-(\cos x-\sin x), \quad \cos x- \sin x<0}\]

Answer 17

At x=90 degrees,\[\Large\bf\sf y(90^o)\quad=\quad |\cos90^o-\sin90^o|\quad=\quad |0-1|\]See how the inside is negative?
That means our absolute function is negative at this particular x coordinate,\[\Large |\cos x-\sin x|=\cases{\quad \cos x-\sin x,\qquad \cos x-\sin x>0\\\color{orangered}{-(\cos x-\sin x), \quad \cos x- \sin x<0}}\]The cosx - sinx gave us something less than 0.
So we use this part of our piece-wise function.

Answer 18

Sorry if this is super confusing :(
Absolute was always a bit weird to me also..
Trying to figure out the right way to word this.

Answer 19

how you thought that cos x - sin x is < 0

Answer 20

\[\Large\bf\sf y(x)=|\color{royalblue}{\cos x-\sin x}|\]The blue part is what tells us what sign to use on our absolute when we want to drop the bars.

So for example:\[\Large\bf\sf y=|x|\]At x=-2,\[\Large\bf\sf y=|-2|\]You might think, "Oh the absolute bars tell us to destroy the negative sign and just give us a positive."
But that's not what's happening.
If we remove the absolute bars we can maybe more clearly see what's going on.
\[\Large\bf\sf y=(-2)\]The absolute is telling us that this needs to be positive.
So we need a negative in front to make it positive,\[\Large\bf\sf y=-(-2)\]
So the example is telling us that at x=-2, our function \(\Large\bf\sf y=|x|\) is actually \(\Large\bf\sf y=-x\)

Answer 21

So to answer your question, ummm...

Answer 22

\[\Large\bf\sf y(x)=|\color{royalblue}{\cos x-\sin x}|\]When the blue stuff is negative, our function is actually:\[\Large\bf\sf y(x)=-(\cos x-\sin x)\]

Answer 23

That's what's getting rid of the negative value inside of the bars.

Answer 24

okay

Answer 25

but why particularly this | cos x - sin x |

Answer 26

Hmm? 0_o

Answer 27

At x=90 degrees, (the stuff in the brackets is negative. So we put a negative in front to compensate. Our function is,\[\Large\bf\sf y(x)\quad=\quad \sin x- \cos x\]I distributed the negative sign.

Answer 28

And it's easier to take a derivative from there.

Answer 29

Thanks sir