√36.6 is near √36 so we will use f(x)=√x with x=36 and ∆x=0.6
Note that f′(x)=‌

1

2√x

√36.6=f(x+∆x)
≈f(x)+f′(x)∆x
=√x+‌

1

2√x

∆x
=√36+‌

1

2√36

0.6
=6.05
OR
The given curve is y=2cos2(3x) .
Differentiating both sides with respect to x , we get:

dy

dx

=−12sin‌(3x)‌cos(3x)
Now, at x=

π

6

, we have:

dy

dx

=−12sin‌(

π

2

)‌cos(

π

2

)=0
This means that the slope of tangent to the curve at x=

π

6

is zero.