Differentiation

1. y=√(1 √x)=(1 x^1/2)^1/2

so dy/dx=1/2(1 √x)^(-1/2)1/2x^(-1/2)

dy/dx=1/[4√(1 √x)(√x)]



2.y=tan(x/2)
dy/dx=[sec^2(x/2)](1/2)
=1/2sec^2(x/2)



3.
y=x^3-4x^2-16x 15

dy/dx=3x^2-8x-16

put dy/dx=0
3x^2-8x-16=0
(3x 4)(x-4)=0
x=-4/3 x=4
(can be skipped
d^2y/dx^2=6x-8

d^2y/dx^2 is smaller than0; x=-4/3
d^2y/dx^2 is larger than 0 ; x=4
so y is max when x=-4/3
y is min when x=4)

so the points are ( -4/3,725/27)
(4,-49)
the y-coordinates are found by substitution

1. dy/dx
=d√(1+√x)/d(1+√x)．d(1+√x)/dx
=1/2√(1+√x)．1/2√x

2. dy/dx
=dtan(x/2)/d(x/2)．d(x/2)/dx
=sec^2(x/2)．(1/2)

3. dy/dx
=4．dcos(sin√x)/d(sin√x)．d(sin√x)/d√x．d√x/dx
=4．-sin(sin√x)．cos√x．1/2√x

4. dy/dx=3x^2-8x-16
when dy/dx=0, (3x+4)(x-4)=0
x=4 or x=-4/3