F (x), G (x) can derive the derivatives of the following functions 1) y = f (x + e to the - x power) 2) y = f (e to the x power) × G (x) power of E 3) y = [XF (x ²)

F (x), G (x) can derive the derivatives of the following functions 1) y = f (x + e to the - x power) 2) y = f (e to the x power) × G (x) power of E 3) y = [XF (x ²)

(1) Y = f (x + e ^ (- x)) y '= f' (x + e ^ (- x)) * (1-e ^ (- x)) (2) y '= f' (e ^ x) * (e ^ x) * e ^ (g (x)) + F (e ^ (x)) * e ^ (g (x)) * g '(x) = = e ^ g (x) (f' (e ^ x) * e ^ g (x) + F (e ^ x) * g '(x)) (3) if y = XF (x ^ 2), then y' = f (x ^ 2) + 2x ^ 2 * f '(x ^ 2)

Given cos2x = 3 / 5, find SiNx ^ 4 + cosx ^ 4

cos2x=3/5
Then sin2x = ± 4 / 5
sinx^4+cosx^4
=(sinx^2+cosx^2)^2-2sinx^2cosx^2
=1-1/2（sin2x）^2
=1-1/2*16/25
=17/25

Given - π / 2 < x < 0, SiNx + cosx = 1 / 5, find cos2x

sinx+cosx=1/5
Square on both sides:
(sinx+cosx)^2=1/25
1+2sinxcosx=1/25
2sinxcosx=-24/25
Using the double angle formula:
sin2x=-24/25
∵-π/2

The range of the function y = SiNx + cosx + sinxcosx, (x ∈ R) is _

Let t = SiNx + cosx=
2sin（x+π
4）∈[-
2，
2] , then T2 = 1 + 2sinxcosx,
Therefore, the function y = SiNx + cosx + sinxcosx = t + t2-1
2=1
2（t+1）2-1，
When t = - 1, the minimum value obtained by the function is - 1=
2, the maximum value obtained by the function is
2+1
2，
Therefore, the value range of the function is [- 1,
2+1
2]，
So the answer is: [- 1,
2+1
2]．

The value range of function y = (SiNx + 1) / cosx + 3 is

In your question, is 3 on the denominator or outside?
I'll solve it for you by pressing the denominator. If it's outside, the method is the same
(sinx+1)/(cosx+3)=y
sinx+1=ycosx+3y
SiNx ycosx = 3y-1 = root sign (1 + y ^ 2) sin (x + P) = 3y-1
Sin (x + P) = (3y-1) / root sign (1 + y ^ 2)
|(3y-1) / root sign (1 + y ^ 2) | = (3y-1) ^ 2
Just solve the inequality
This problem can also be solved by combining the slope and number shape of points (cosx, SiNx) and points (- 3, - 2)

Find the value range of function y = cosx (cosx + SiNx) Such as title~

y=cosx(cosx+sinx)
=cos ² x+sinxcosx
=(cos2x+1)/2+1/2·sin2x
=1/2·(sin2x+cos2x)+1/2
=1/2·√2(√2/2·sin2x+√2/2·cos2x)+1/2
=1/2·√2(cosπ/4·sin2x+sinπ/4·cos2x)+1/2
=1/2·√2sin(2x+π/4)+1/2
=√2/2·sin(2x+π/4)+1/2
∵sin(2x+π/4)∈【-1,1】
∴1/2-√2/2≤√2/2·sin(2x+π/4)+1/2≤1/2+√2/2
Therefore, the value range is: [1 / 2 - √ 2 / 2,1 / 2 + √ 2 / 2]