What is the derivative of e to the - x power?

What is the derivative of e to the - x power?

Equal to the - x power of - e

Ask the master advanced mathematics question: y = (upper limit of definite integral 2 π lower limit 0 subject: the square power DT of T of E), what is the derivative of y to x?

0 (there is no X in the definite integral, so the result of the definite integral is a constant independent of x)

Known function f (x) = sin (x)+ θ）+ cos（x- θ） Is an even function, then θ The value is __

∵f（-x）=sin（-x+ θ）+ cos（-x- θ）= sin θ cosx-cos θ sinx+cosxcos θ- sinxsin θ= f（x）=sinxcos θ+ cosxsin θ+ cosxcos θ+ sinxsin θ ∴-cos θ sinx-sinxsin θ= f（x）=sinxcos θ+ sinxsin θ ∴-2sinxcos θ= 2sinxsin θ ∴sin...

Let f (x) = sin (Wx)+ φ)+ cos(wx+ φ) (w>0,| φ|

f(x)=sin(wx+ φ)+ cos(wx+ φ)
=√2sin(wx+ φ+ π/4)
T=2π/w=π
w=2
f(x)=√2sin(2x+ φ+ π/4)
f(-x)=f(x),
So f (- π / 8) = f (π / 8)
sin φ= sin(π/2+ φ)= cos φ
tan φ= one
| φ|

It is known that the minimum positive period of the square Wx (W greater than 0) of the function f (x) = sin (π - Wx) cos Wx + COS is π Find the value of W Shorten the abscissa of each point on the image of function y = f (x) to half of the original, and keep the ordinate unchanged to obtain the image of function y = g (x). Find the minimum value of function g (x) in the interval [0,16 π]

f(x)=sin(π-wx)coswx+cos ² wx
=sinwxcoswx+cos ² wx
=(1/2)sin2wx+(1/2)cos2wx+1/2
=(√2/2)sin(2wx+π/4)+1/2
∵ f (x) has a minimum positive period of π
∴2π/2w=π
w=1
∴f(x)=(√2/2)sin(2x+π/4)+1/2
g(x)=(√2/2)sin(2*2x+π/4)+1/2
=(√2/2)sin(4x+π/4)+1/2
0≤x≤π/16
π/4≤4x+π/4≤π/2
1≤g(x)≤(√2/2)+1/2

If the function f (x) = sin (Wx + AFA) + cos (Wx + AFA) (W > 0, the absolute value AFA

F (x) = sin (Wx + AFA) + cos (Wx + AFA) = √ 2 [√ 2 / 2 sin (Wx + AFA) + √ 2 / 2 cos (Wx + AFA)] (extract √ 2, root 2) = √ 2 [cos45 ° sin (Wx + AFA) + sin45 ° cos (Wx + AFA)] = √ 2 sin (Wx + AFA + 45 °) = √ 2 cos (Wx + AFA - 45 °) (because