√ how to find the derivative under the root sign (square of 1 + x)

√ how to find the derivative under the root sign (square of 1 + x)

This is the derivation of a composite function:
Let y = 1 + x ^ 2, then the original function is √ y
√ the derivative of Y is 1 / 2Y ^ (- 1 / 2)
The derivative of 1 + x ^ 2 is 2x
The derivative of the original function is 1 / 2Y ^ (- 1 / 2) · (2x) = 1 / 2 (1 + x ^ 2) ^ (- 1 / 2) · (2x)
Then put it in order: X / (√) (1 + x ^ 2)

Given SiNx + cosx = 2 / 2 under the root sign, find the values of sinxcosx, SiNx cosx, sin ^ 2x cos ^ 2x, sin ^ 4x + cos ^ 4x, sin ^ 6x + cos ^ 6x

The answers are: negative 1 / 4, 3 / 2, radical 15 / 4

F (x) = (1 + SiNx + cosx) [sin (x / 2) - cos (x / 2)] / root sign (2 + 2cosx) (1) Simplified form of F (x) (preferably with process)

f(x)=(1+sinx+cosx)[sin(x/2)-cos(x/2)]/√（2+2cosx）
=[sin ² (x/2)+cos ² (x/2)+sinx+cosx][sin(x/2)-cos(x/2)]/[√2（1+cosx）]
=[sin ² (x/2)+cos ² (x/2)+2sin(x/2)*cos(x/2)+cos ² (x/2)-sin ² (x/2)][sin(x/2)-cos(x/2)]/[√4*（1+cosx）/2]
=｛[sin(x/2)+cos(x/2)] ²+ [cos(x/2)+sin(x/2)][cos(x/2)-sin(x/2)]｝[sin(x/2)-cos(x/2)]/[√4*cos ² (x/2)]
=｛[sin(x/2)+cos(x/2)][sin(x/2)+cos(x/2）+cos(x/2)-sin(x/2)]｝[sin(x/2)-cos(x/2)]/[√4*cos ² (x/2)]
=｛[sin(x/2)+cos(x/2)]*2cos(x/2）｝[sin(x/2)-cos(x/2)]/[2*cos(x/2)]
=｛[sin(x/2)+cos(x/2)][sin(x/2)-cos(x/2)]*2cos(x/2）｝/[2*cos(x/2)]
=[sin(x/2)+cos(x/2)][sin(x/2)-cos(x/2)]
=sin ² (x/2)-cos ² (x/2)
What I don't understand

(COS (x / 2) / (root sign 1 + SiNx)) + (sin (x / 2) / root sign (1-cosx)), X belongs to (3 / 2 π, 2 π)

∵sin(x/2)=±√[(1-cosx)/2]
cos(x/2)=±√[(1+cosx)/2
∵ x ∈ (3 π / 2,2 π)
∴3π/4

Given SiNx · sin under the root sign ² X + cosx · cos under the root sign ² X = - 1, then the quadrant of X is

SiNx · sin under the root sign ² X + cosx · cos under the root sign ² X = SiNx ▏ SiNx ▏ + cosx ▏ cosx ▏ = - 1 indicates that both SiNx and cosx are negative, so x is the third quadrant

SiNx cosx = 1 / 2 find the value of sin * 3x cos * 3x?

SiNx cosx = 1 / 2, the square can get sinxcosx = 3 / 8. Sin ^ 3x cos ^ 3x = (SiNx cosx) (sin ^ 2x + sinxcosx + cos ^ 2x) = 1 / 2 * (1 + 3 / 8) 11 / 16