Derivation: x = cos ^ 4 * t, y = sin ^ 4 * t, dy / DX

Derivation: x = cos ^ 4 * t, y = sin ^ 4 * t, dy / DX

DX / dt = 4 (cost) ^ 3 * (cost)'dy / dt = 4 (SIN) ^ 3 * (SIN)'and (cost) '= - Sint (SIN)' = cost, so dy / DX = (dy / DT) / (DX / DT) = 4 (cost) ^ 3 * (- Sint) / [4 (SIN) ^ 3 * cost] = - (COT T) ^ 2

y=arctanx y=arctanx Derivative y’=1/tany y’=1/x Is this how to seek derivation? I also forgot the answer I don't know how to explain

The first result is 1 / (1 + x ^ 2)
Derivation process
x=tany
Deriving x
1=y'*sec^2y
=>y'=1/sec^2y=1/(tan^2y+1)=1/(x^2+1）
If you feel good, please accept it. If you don't understand, you can ask

Derivative: y = ln ^ 2 (1 + x)

y‘=2ln(1+x)/(x+1)

Find the derivative, y = ln (x + √ x ^ 2 + 1),

Derivative method of compound function
Let u = x + √ (x ^ 2 + 1), then y = LNU
y'=(lnu)'=(1/u)*(u)'=(1/u)*(1+x/(√x^2+1))=(x+√(x^2+1))*(1+x/(√x^2+1))

Derivative! Y = ln √ (1 + 2x)

y=ln(2x+1)^1/2
=1/2*ln(2x+1)
So y '= 1 / 2 * 1 / (2x + 1) * (2x + 1)'
=1/2*1/(2x+1)*2
=1/(2x+1)

For the function y = ln (3x), take the derivative If y = Ln3 + LNX, then we can get 1 / X. in my question, we can't take 3x as a whole and get 1 / 3x?

Derivative of composite function: F (g) '= f' (g) * g '
So y '= (ln3x)' * (3x) '
=1/3x *3
=1/x

Y = (x √ x + 3) e ^ 2x derivative

y=[x^(3/2)+3]e^2x
y`=(3/2√x)e^2x+2[x^(3/2)+3]e^2x
=e^2x {3/2√x+2x^(3/2)+6}

X of 3 plus 1 times of 2 x + 1 power = 6 of 2x-3 power solution equation

3^(x+1)x2^(x+1)=6^(x+1)=6^(2x-3),
x+1=2x-3,x=4

Calculation question: (- 0.25) 2006 power × (- 4) 2008 power × (- 1) 2009 power The power of 2006, power of 2008 and power of 2009 are in the upper right corner of bracket. I am so anxious!

Equal to - 16

The cubic power of 2009 - 2 × 2009 square - 2007 △ 2009 cubic + 2009 square - 2010

(2009³-2×2009²-2007) ÷( 2009³+2009²-2010)
=[2009²(2009-2)-2007]/[2009²(2009+1)-2010]
=(2009²×2007-2007]([2009²×2010-2010)
=[(2009²-1)×2007]/[(2009²-1)×2010]
=2007/2010=(2010-3)/2010=1-(3/2010)
=1-0.00149=0.98851