Question

Not sure where to start... find the derivative of y = sin(sinx + cosx). Can you mention rules used please?

Answer 1

\(y = \sin(\sin(x) + cos(x))\), right?

Answer 2

yep that's it!

Answer 3

We can use chain rule here very well.. :)

Answer 4

\[y' = \frac{d}{dx}\sin(\sin(x) + \cos(x)) \times \frac{d}{dx}(\sin(x) + \cos(x))\]

Answer 5

Firstly, we have to take derivative of whole \(sin(w)\), then we do the derivative of \(w\), where \(w = sin(x) + cos(x)\)

Answer 6

Getting?

Answer 7

good so far! (ignore when it says I'm typing, my comp is effed up haha)

Answer 8

The same it does with me, so don't take it granted that when you see that, then it means I am typing only.. :P

Answer 9

So, what is the derivative of \(\sin(x)\), can you tell?

Answer 10

cosx, right? so would the whole derivative be... (cos(cosx-sinx))(cosx-sinx) ?

Answer 11

Wait.. ")

Answer 12

Don't go with the speed of light, we will check it one by one, alas.!! you are not right with the whole derivative.. :( But you are right about cos(x).. :)

Answer 13

haha okay, I figured it wouldn't be that simple ):

Answer 14

This way the derivative of \(sin(w)\) will be \(cos(w)\), right?

Answer 15

So, we are now having:

\[y' = \cos(\sin(x) + \cos(x)) \times \frac{d}{dx}(\sin(x) + \cos(x))\]

Answer 16

Getting?

Answer 17

we will firstly differentiate sin only, but not its angle, for angle we have second step.. :)

Answer 18

mmkay, so even though sinx's derive is cos, the values inside (w) will remain the same even if they are sin and cos?

Answer 19

They are angle or you can say they are argument, see:
\[\sin(\frac{\pi}{3})\]
Then agnle pi/3 is just same as (sin(x)+cos(x) there..

Answer 20

Remember this: first the function, then its angle.. For trigonometric case.. :)

Answer 21

ahh alright, I see!

Answer 22

*angle not agnle, there.. :)

Answer 23

So, can we proceed further?

Answer 24

yes, I'm ready!

Answer 25

\[\frac{d}{dx}(\sin(x) + \cos(x)) = \frac{d}{dx} \sin(x) \times \frac{d}{dx}(x) + \frac{d}{dx}(\cos(x)) \times \frac{d}{dx}(x)\]

Answer 26

This is for the second part, see first the function and then its angle.. Getting?

Answer 27

hmm, okay. I think so, but we're also deriving the (x)'s? that will just be 1, right?

Answer 28

Yep, that is why sin(x) derivative is cos(x):

It is actually like this:
\[\frac{d}{dx}(sin(x)) = cos(x) \times \frac{d}{dx}(x) = cos(x) \quad only.\]

Answer 29

I am just showing it to prove that we differentiate function like sin, cos etc then we take derivative of its angle.. :)

Answer 30

oh okay, I get it then!

Answer 31

So, can you tell now what is our whole derivative?

Answer 32

(sorry, my computer hates this website haha)
y1 = (cos(sinx+cosx))(cosx-sinx)?

Answer 33

Superb.. :)

Answer 34

yay thank you so much for all your help!

Answer 35

It will be more nice if you bring out (cos(x)-sin(x)) in the front..

Answer 36

\[\color{blue}{y' = (\cos(x)-\sin(x)) \cdot \cos(\sin(x) + \cos(x))}\]

Answer 37

In multiplication, you can change the order which you are writing in.. :)

Answer 38

ahh I see, excellent! thanks again!! (:

Answer 39

\(\text{This is it}...\)

you are welcome dear.. :)