Why is the derivative of Sin & # 178; X sin2x? It should not be brought in according to the formula X & # 178; = 2x ^ X-1, but that would be wrong

Why is the derivative of Sin & # 178; X sin2x? It should not be brought in according to the formula X & # 178; = 2x ^ X-1, but that would be wrong

The formula is X & # 178; = 2x
sin²x=2sinxcosx=sin2x

How to calculate the derivative of function y = xtanx

y'=x'tanx+x(tanx)'
=tanx+x*scs^2(x)
=tanx+x/cos^2(x)

Derivation: y = (sin2x + (TaNx) ^ 2) ^ 1 / 2

y'=1/2*(sin2x+(tanx)^2)^(-1/2)*[2cos2x+2tanx*(secx)^2]
And then to simplify

Derivation: y = TaNx

y'=sec^2x
Or y '= (SiNx / cosx)' = (Co ^ 2x + sin ^ 2x) / cos ^ 2x = sec ^ 2x

If f (sin (x / 2)) = 1 + cosx, then the n-th derivative of F (COS (x / 2)) is obtained=
The n-th derivative of F (COS (x / 2)) is (d ^ NF (COS (x / 2))) / (DX ^ n)

F (sin (x / 2)) = 1 + cosx = 1 + 1-2sin ^ 2 (x / 2) = 2-2sin ^ 2 (x / 2) f (x) = 2-2x ^ 2F (COS (x / 2)) = 2-2cos ^ 2 (x / 2) = 1-cosxf '= sinxf' '= cosxf' '= - sinxf' '= - cosxf ^ (5) = - sinxf ^ (6) = - cosxf ^ (7) = SiNx six cycles can be seen

How to find the derivative of y = sin ^ 4x / 4 + cos ^ 4x?
Wrong number, y = sin ^ 4 (x / 4) + cos ^ 4 (x / 4)

y=sin^4(x/4)+cos^4(x/4)
=[sin^2(x/4)+cos^2(x/4)]^2-2sin^2(x/4)cos^2(x/4)
=1-1/2sin^2(x/2)
=1-1/4(1-cosx)
=3/4+1/4cosx
y′=-1/4sinx

Find the derivative of y = sin ^ 5x, y = cos (4x + 3),

If y = g (U), u = f (x), then dy / DX = dy / Du * Du / DXY = sin ^ 5xdy / DX = DSIn ^ 5x / dsinx * dsinx / DX = 5sin &; xcosxy = cos (4x + 3) dy / DX = DCOS (4x + 3) / D (4x + 3) * D (4x + 3) / DX = - sin (4x + 3) * 4 = - 4sin (4x + 3)

The image of the function y = 5sin (2x - π / 6) + 1 can be obtained from the image of the function y = SiN x after oxygen enrichment
Please write clearly 233

First, we move π / 12 along the x-axis,
Then it shrinks to 1 / 2 with the origin as the center in the X direction;
Then, it is magnified 5 times along the y-axis;
Finally, move 1 unit along the y-axis

What is the monotone increasing interval of the square X of the function y = cos x-sin?

Y = cos square x-sin square x
=cos²x-sin²x
=cos2x
The monotone increasing region of y = cos2x is:
π + 2K π ≤ 2x ≤ 2 π + 2K π, K is an integer
That is, π / 2 + K π ≤ x ≤ π + K π, K is an integer
The monotone increasing interval of the original function is [π / 2 + K π, π + K π], K is an integer

Monotone increasing interval of function f (x) = cos square x-sin square x

Let's change it to cos (2x) increasing range from KPI + pi / 4 to KPI + 3pi / 4