>sin 2x + 1
Is this either:
>sin(2x) + 1
or
>sin(2x + 1)
>>7888840
it's sin(2)*x+1
>>7888840
Could be either. Check your book for answer
>>7888840
s*i*n*2*x + 1
>>7888843
kek
>>7888852
you killed it
>>7888858
havent heard this in years
>>7888840
>>7888840
If there are no parenthesis then assume sin(2x) +1
[math]d/dx sin(2x) + 1 = 2cos(2x)[/math]
>>7888872
>failing a basic derivative
incredible, you even forgot the constant
d/dx sin(2x) + 1 = 1 + cos(2x)
>>7888878
I can't believe people like you exist.
>>7888878
r u retarded?
>>7888872
>not [math]\frac{d}{dx}\sin(2x)+1=2\cos(2x)[/math]
>>7888878
>d/dx sin(2x) + 1 = 1 + cos(2x)
Same shit as [math]\frac{d}{dx} sin(2x)+1 = 2cos(2x)[/math]
>>7889042
>[math]\frac{d}{dx} \sin(2x)+1 = 2\cos(2x)[/math]
It's actually [math]2\cos(2x) + 1[/math]
>>7889459
The derivative of a constant is zero
>>7889459
b-but senpai... ;__;
>>7889923
ddx(sin(2x))+1=2cos(2x)+1 for all x.
>>7889939
the entirety of sin(2x) + 1 is being derived with respect to x
>>7889919
pls no bully
>>7889459
are you deriving the sin and keeping the constant?
fuck use brackets or just derive the constant, which equals 0
i cant believe i replied to this
>>7890966
Ignore them
>>7889919
>Differential operator distributes, tard.
Grown ups say that it's a linear operator.
>this whole thread
you're all dinguses and I suggest imminent sudoku
>>7892864
>LEARN OPERATORS.
No! c:
I would read it as [math]\sin(2) \, x \,+\, 1[/math]. It should be written without brackets only when there is only one symbol in it.
>>7891469
I just derived/differitiated that using limits.
I have no fucking idea how they got to keep the constant. It is 2cos(2x)
>>7893115
nah, it is sin(2x)
If x is not part of the function write it as a coefficient meaning it comes first but then again I guess it depends on the country.
>>7888840
Stop making this thread.
>>7892864
f(x) = sin(2x) + 1
f ' (x) = 2cos(2x) + 0
So, chicken=[math]e^{-x}[/math] and egg=[math]-e^{-x}[/math]?
