Function f (x) = cos2x − 1 The period of 2 is   （　　） A. π four B. π two C. 2π D. π

Function f (x) = cos2x − 1 The period of 2 is   （　　） A. π four B. π two C. 2π D. π

f(x)＝cos2x−1
two
=1
2（2cos2x-1）
=1
2cos2x，
The period is t = 2 π
2=π
Therefore: D

Y = cos ^ 2 (x ^ 2 + 1) derivative, what method do you use

y'=[cos^2(x^2+1)]'
=[2cos(x^2+1)]·[cos(x^2+1)]'
=[2cos(x^2+1)]·[-sin(x^2+1)]·[x^2+1]'
=[2cos(x^2+1)]·[-sin(x^2+1)]·2x
The middle uses the chain derivation law

Can both sides of an identity take derivatives and maintain equality identity at the same time? Why? For example, x ^ 2 + y * x + 1 = 0

Identity means that both sides represent the same function. Their derivatives are also the same (a function has only one)
It's a derivative function), so after the derivatives on both sides, it's still an identity
X ^ 2 + y * x + 1 ≡ 0 ≡ derivation on both sides, 2x + y + XY ≡ 0 (x is the independent variable. Y is the dependent variable)
This is a common method for deriving implicit functions

Take the derivative of X on both sides of the equation, XY + y-x-8 = 0. What does this mean, ask an expert for advice Take the derivative of X on both sides of the equation, XY + y-x-8 = 0. What does this mean? Ask an expert for guidance and explain it clearly

This topic involves the expression of implicit function derivation:
1. This is a definition method, which defines that y is a function of X through the equation XY + Y - X - 8 = 0,
This kind of equation can not solve a specific expression, y = f (x)
If you can write an expression like y = f (x), it is called that y is an explicit function of X;
If you can't write a specific expression, it is called that y is the implicit function of X
2. Whether explicit function or implicit function; Whether y = f (x) is solved or not, it cannot change that y is X
The fact of a function
So XY + Y - X - 8 = 0, both sides are functions of X, and both sides take derivatives of X,
The left becomes: y + XY '+ y' - 1 (indicating that the left is both a function of X, a function of Y, and finally a function of X,
X takes the derivative of X, y also takes the derivative of X, which is the chain derivative of implicit function);
The right becomes: 0 (indicating that the right does not contain X and y, and is not a function of X and y)
So, y + XY '+ y' - 1 = 0, y '= (1 - y) / (1 + x)
Similarly, the expression of Y 'is not necessarily x, but may also include y. that's how the derivative of the implicit function is expressed
It may not be solved, it may be unwilling to solve, there is no need to solve, it doesn't matter, there are no regulations, and it must be expressed explicitly

Under what circumstances can an equation be derived from both sides?

Both derivative functions exist in the definition domain, that is, both sides can be derived in the definition domain

On both sides of the original equation, find the indefinite integral. Is the equation still valid? On both sides of the original equation, find the derivative. Is the equation still valid?

The two sides of the original equation, after solving the indefinite integral, the equation is not tenable, because after solving the indefinite integral, an arbitrary constant will be added. The two sides of the original equation are tenable after the derivative