How do you reduce this function: y′=2sin(x)[cos(cos(x))sin(cos(x))−cos(cos(x))sin(cos(x))]
#Trig Function reduction
velvet cradle
trail flickerBOT
dire anchor
How do you reduce this function: y′=2sin(x)[cos(cos(x))sin(cos(x))−cos(cos(x))si...
Isn't cos(cos(x)) • sin(cos(x)) − cos(cos(x)) • sin(cos(x)) = 0
velvet cradle
Wouldn't that be equalivent to for example 2sin(x)a-a? And doesn't follow PEMDAS right?
velvet cradle
So substituting some random variables 2x*5-5?
velvet cradle
2sin(x)[a-a] is the equivalent
Is this an alegbra rule?
If so, can you let me know which one.
dire anchor
distributive property
a(b+c) = ab + ac
2sin(x)[a-a] = 2sin(x)[0] = 0
or
2sin(x)[a-a] = 2sin(x)a - 2sin(x)a = 0
velvet cradle
Okay. Let me try to rework this problem real quick
dire anchor
I don't understand your problem anyway
velvet cradle
The original question was find y': y=cos^2cos(x)+sin^2(cosx)
arctic yarrowBOT
adonhs
dire anchor
The original question was find y': y=cos^2cos(x)+sin^2(cosx)
That's funny, you don't need chain rule 😄
arctic yarrowBOT
adonhs
dire anchor
So actuall you have y = 1 and the derivative of a constant is always 0
velvet cradle
Yes, I'm familiar with that identify. But I wanted more practice with the chain rule using prime notation
dire anchor
Oh that's fine
arctic yarrowBOT
adonhs
dire anchor
Yeah I think you did very well
velvet cradle
I had help with the answer I got, actually. Can you help walk me thru the problem?
dire anchor
which problem
velvet cradle
The problem we just solved
dire anchor
well it's essentially chain rul
velvet cradle
Can I walk thru the steps?
arctic yarrowBOT
adonhs
dire anchor
Same with 1 more funciton
Keep in mind you always keep the argument
And chain rule works like, first you differentiate the outer function, then you differentiate the inner ones
velvet cradle
So the first expansion is $$\frac{d}{dx}(cos(cosx))^2$$
arctic yarrowBOT
jboi72
velvet cradle
Then $$2(cos(cosx)\frac{d}{dx}(cos(cosx))\frac{d}{dx}(cosx)$$
arctic yarrowBOT
jboi72
arctic yarrowBOT
adonhs
dire anchor
This only
arctic yarrowBOT
adonhs
dire anchor
**jboi72**
Otherwise you differentiate here cos(x) twice
dire anchor
**adonhs**
the derivative of cos(x) is already here
One chain rule has basically 2 derivatives
velvet cradle
Okay. and the we get $$2sinx(cos(cosx)sin(cosx)$$
dire anchor
You can basically imagine that the differential operator d/dx is moving in to the next inner function kinda
after getting down with the previous derivative
velvet cradle
So d/dx for $$y=cos^2(cos(x))=2(cos(cosx)-sin(cosx)-sinx$$
arctic yarrowBOT
jboi72
dire anchor
i can't follow
velvet cradle
d/dxcos(x) = -sin x correct?
dire anchor
yes
velvet cradle
**adonhs**
Wasn't this your answer?
velvet cradle
**jboi72**
So this is the first part of our equation, correct?
dire anchor
I am sorry but the way you wrote it confuses me
dire anchor
**adonhs**
I thought this would be self-explanatory
velvet cradle
I thought this would be self-explanatory
We next need to find the derivative for $$sin^2x(cosx)$$
, correct?
arctic yarrowBOT
jboi72
dire anchor
Yeah also
I just did the cosine part
it's about the principle
also it would be double the work to write the whole shit🫠
velvet cradle
$$-2sin(cosx)(cos(cosx)sinx$$
arctic yarrowBOT
jboi72
velvet cradle
This is the second answer correct?
velvet cradle
so place -2sinx in front and then factor it out?
dire anchor
yea factor out 2sin(x)
the minus is not a common factor tho
but if you still factor it out you need to make up for it
like
2ax - 2ay = 2a(x-y)
your way:
2ax - 2ay = -2a(-x+y) = -2a(y-x)
velvet cradle
Okay. Yeah, so I did it right. Initially, just didn't factor out.
Anything you want to add, if not I'll close this post.
dire anchor
nah man haha