If f (x) = sin (x + a) + cos (x-a) (0

If f (x) = sin (x + a) + cos (x-a) (0

According to the definition of even function, f (x) = sin (x + a) + cos (x-a) = sinxcosa + cosxsina + cosxcosa + sinxsina = SiNx (COSA + Sina) + cosx (Sina + COSA) SiNx is an odd function and cosx is an even function. Therefore, if f (x) is an even function, the coefficient of SiNx is 0, that is, cosa + Sina = 0 and 0

Design | a| The answer is a = 0.75 π

F（x）=sin(x+a)+cos(x-a)
F(-x)=sin(a-x)+cos(a+x)
F (x) + F (- x) = root 2Sin (x + A + 0.24 π) + root 2Sin (A-X + 0.25 π) = 0
So sin (x + A + 0.25 π) + sin (A-X + 0.25 π) = 0
So you have to make one SiNx and the other - SiNx
Therefore, it is easy to know that a = 0.75 π
That just changed the cosine into sine

What function is y = x ^ x (y = x to the x power)? What is the definition field and value field? What is the derivative? How exactly? How to draw the image of this function? Can you expand it into a McLaughlin function (expand 4 terms). What are non differentiable points and discontinuous points? What are the function values and derivative values at x = 0?

Y = x ^ x is neither a power function nor an exponential function. The definition field (0, + ∞), the value field [1 / e ^ (1 / E), + ∞), and the derivative is x ^ x (LNX + 1). The specific number of words is limited and cannot be written
It cannot be expanded into McLaughlin because x = 0 is not in the definition domain. There are no non differentiable points and discontinuous points. F (0) = + ∞, f '(0) = - ∞

Derivative of exponential function y = a ^ x, change amount △ y = x + △ x power of a - A ^ x = a ^ x (Δ x power of a - 1), △ Y / △ x = a ^ x (A's △ x power - 1) / △ x = a ^ x multiplied by (A's △ x power - 1) / △ x, take the limit: let u = A's △ x power - 1, then △ x = loga (1 + U). How did this step come about? Log this step?

△ x power of u = a - 1
△ x power of a = u + 1
Take logarithms from both sides
LG (the △ x power of a) = LG (U + 1)
△xlga=lg(u+1)
△x=lg(u+1)/lga
△x=loga（1+u）

The derivative of the X / K power of E is

Guo Dunyong replied:
F(x)=e^x/k
F(x)′=e^x/k × (x/k)′=(e^x/k)/k

What is the derivative of the X-1 power of E? rt

(x2+2x)e^x-1
Hehe, the answer to the whole question~