Does y differentiate into:(y)'=1y^0y'=1 or (y)'=yy'?
Update:
A sample problem for this equation would be
Sin^2(x)-y=(Constant)
solving through implicit differentiation i get
2Sin(x)*Cos(x)-y'=0
2sin(x)*Cos(x)=y'
or
2Sin(x)*Cos(x)-yy'=0
-(2Sinc(x)*Cos(x))/(-y)=y'
The questions also ask to represent the equations using a vector field but that only works if y' is a function of x and y and not just x.
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y' equals y'
Your first answer is correct.
y' = 2sin(x)*cos(x)
You could rewrite that as y' = sin(2x) using a trig identity.
The notation used throughout your question is a bit cumbersome, and has taken a while to comprehend. But I wanted to take a stab at it nonetheless:
Differentiating any function, in this case y, yields the expression y' (read "y prime").
Perhaps you are asking "What if y represents an implicit function?" Implicit differentiation looks different ... so if that is your topic, I recommend you re-phrase the question.
In short, there is normally no need to consider "an inside function" nor to apply the well-known Chain Rule. Are you asking about the Chain Rule?
Essentially, the derivative of y(x) is y'(x). But if the inside argument is more "complicated" than just x, you might write:
dy/dx [y(f(x))] = y'(f(x)) f'(x) [which is Chain Rule]
Hope this helps!
That depends, what are you differentiating with respect to, and what is the function you are differentiating?