Differential Calculus

I have a problem sir.

h=1/2√4a^2-(b-a)^2

A=1/4(b+a)√4a^2-(b-a)^2

dA/db=1/4[(b+a)-2(b-a)/2√4a^2-(b-a)^2 + √4a^2-(b-a)^2]=0.

Question:
How did it become +√4a^2-(b-a)^2]=0?
How did it become zero? Why is there zero?

Please do leave a solution so I can study it, thank you!

Obtain the differential equation by eliminating the arbitrary constants.
1. Cxsiny+x^(2)y=Cy
2. y=Ae^(-3x)-Be^(2x)+Cx^3
3. Ay=e^(Bx^2)

Hello everyone! I need some help again. Can anyone derive this problem with steps?

d/dx ((tan2x)+(sec2x))
Thank you.

May answer key po ba kayo nito. Icheck ko lang po sana kung tama answers ko. Thank you po. Critical points/maxima and minima

y=4-6x+x²
y=(2x-1)²
y=2+12x-x³
y=x³-3x²-9x+20
y=x³-3x²+4x+5
y=x³-6x²+12x
y=x⁴+2x²+8x+3
y=16x+4x²-x⁴
y=x²(x-2)²
a³y=x⁴
9a³y=x(4a-x)³
a³y=x²(2a²-x²)
a³y=x³(4a-3x)
a⁵y=x⁴(3a²-2x²)

Make the curve y=ax³+bx²+cx+d have a critical point at (0,-2) and also be a tangent to the line 3x+y+3=0 at (-1,0).
Ans nya po is y=x³+3x²-2. Salamat po

What is the condition that the cubic y=ax³+bx²+cx+d shall have two Extremes? Answer nya po is b²-3ac>0 Salamat po.