The derivative of the function y = sin2x - cos2x is () A. 2 2cos(2x-π 4) B. cos2x-sin2x C. sin2x+cos2x D. 2 2cos(2x+π 4)

The derivative of the function y = sin2x - cos2x is () A. 2 2cos(2x-π 4) B. cos2x-sin2x C. sin2x+cos2x D. 2 2cos(2x+π 4)

y′=（sin2x）′-（cos2x）′=2cos2x+2sin2x=2
2（
Two
2cos2x+
Two
2sin2x）=2
2cos(2x-π
4)
Therefore, a

The following functions are derived to the x power of y = (x / 1 + x); the 1 / x power of y = x times the x power of E; and the cos2x power of y = e times the square of sin X What does this symbol mean in mathematics?

The derivation method of these functions is to take the logarithm of Y, that is, LNY =., and then restore it back, where! Is the factorial, for example, 4! = 4 × 3 × 2 × 1
For example, y = (x / 1 + x) ^ x, then LNY = ln (x / 1 + x) ^ x = XLN (x / 1 + x)
Both sides are deriving x at the same time, y '/ y = ln (x / 1 + x) + X × (1 + X / x) × (x / 1 + x)'
=ln(x/1+x)+(1+x)×(1+x)^-2
=ln(x/1+x)+1/(1+x)
So y '= y [ln (x / 1 + x) + 1 / (1 + x)] = (x / 1 + x) ^ x × [ln (x / 1 + x) + 1 / (1 + x)]

Derivative y = 4 / (x + cos2x) to the power of 2

y=4/(x+cos2x)^2
So we can get the derivative
y'=(-2) *4/(x+cos2x)^3 *(x+cos2x)'
obviously
(x+cos2x)'=1 -2sin2x
So the derivative of Y is zero
y'= -8(1 -2sin2x) /(x+cos2x)^3

Find the third power of derivative y = 5x + the power of 2x - the power of 3E

Y = C (C is a constant) y '= 0 2 power function. Y = x ^ n, y' = NX ^ (n-1) (n ∈ Q *) 3. (1) y = a ^ x, y '= a ^ xlna; (2) remember y = e ^ x, y' = e ^ x, the only derivative function is its own function 4. (1) y = logax, y '= 1 / xlna (a > 0 and a is not equal to 1, x > 0)

Where is the point where the derivative of this function equals zero?

y'=3x²
Y '= 0, x = 0
So, the point where the derivative is zero is (0,0)

Find the derivative of function y = (1 / 3) to the power of X by definition

It is noticed that: (1 / 3) ^ H-1 = e ^ (- hln3) - 1 is equivalent to - hln3y = (1 / 3) ^ XY '= Lim [h → 0] [f (x + H) - f (x)] / h = Lim [h → 0] [1 / 3) ^ (x + H) [(1 / 3) ^ x] / h = Lim [h → 0] (1 / 3) ^ x [(1 / 3) ^ H-1] / h = Lim [h → 0] (1 / 3) ^ x [e ^ (- hln3) - 1] / h = Lim [h → 0] (1 / 3) ^ x [e ^ (- hln3) - 1] / h = Lim [h → 0] (1 / 0] (1 / 1 / 0) (1 / 0] (1 / 3 / 3) (1 / 3) - 3) ^ x (- hln3)

Find the derivative of x power * 2 of function y = 3

The x power of y = 3 * the x power of 2 = the x power of 6
Then y '= 6 to the power of x times ln6

Derivation of X to the power of X

x^x = e^(xlnx)
So (x ^ x) '= [e ^ (xlnx)]' = e ^ (xlnx) * (xlnx) '= x ^ x * (x * 1 / x + LNX) = x ^ x * (1 + LNX)

To the power of X of derivative y = (1 + 1 / x)

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Derivation y = (1 + X (2nd power)) (x power)

This paper uses logarithmic derivation method: LNY = ln (1 + x ^ 2) ^ xlny = XLN (1 + x ^ 2) ^ xlny = XLN (1 + x ^ 2) two sides of the X derivation: y '/ y = ln (1 + x ^ 2) + X / (1 + x ^ 2) * (1 + x ^ 2)' = ln (1 + x ^ 2) + 2x ^ 2 / (1 + x ^ 2) so y / = (1 + x ^ 2) ^ x * [ln (1 + x ^ 2) + 2x ^ 2 / (1 + x ^ 2)] Note: because x is the Y is y, because x is y is the y of Y, because x is the Y is y, because x is the Y is the Y is the y of Y, because x is the X is the Y is the Y is the y of y is the Y is the function, so y is going to find X