Find the derivative of the function. Simplify if possible. and...... y = 17 arctan(sqrt x) y = arcsin(4x + 2) y = arccos(e8x) h(t)...

1) To calculate the derivative of y = sqrt(3arctan
x),we'll apply the chain rule:

y' = [sqrt(3arctan
x)]'*(3arctan x)'

We'll see 3arctan x as an entity and
we'll differentiate the sqrt:

(sqrt t)' = 1/2sqrt
t

y' = [1/2sqrt(3arctan
x)]*[3/(1+x^2)]

y' =
3/[2*(1+x^2)*sqrt(3arctan x)]

2) We'll
calculate the derivative of  y = 17 arctan (sqrt x) using the chain rule
also:

y' = [17 arctan (sqrt x)]'*(sqrt
x)'

y' =
[17/(1+(sqrtx)^2)]*(1/2sqrtx)

y' =
17/2(sqrtx)*(1+x)

3) We'll calculate the
derivative of  y = arcsin(4x + 2)using the chain rule
also:

y = [arcsin(4x
+ 2)]'*(4x+2)'

We'll see 4x + 2 as an
entity:

y' =
{1/sqrt[1-(4x+2)^2]}*(4)

We'll expand the
square:

y' = 4/sqrt(1-16x^2 - 8x -
4)

We'll combine like
terms:

y' = 4/sqrt(-3-16x^2 -
8x)

4) We'll calculate the derivative of  y
= arccos(e^8x) using the chain rule
also:

y =
arccos(e^8x)

We'll
put (e^8x) =
t

(arccos t)' =
1/sqrt(1-t^2)

y' =
[arccos(e^8x)]'*(e^8x)*(8x)'

y'
=
[1/sqrt(1-(e^8x)^2)]*(e^8x)*(8)

y'
=
8(e^8x)/[sqrt(1-(e^8x)^2]

5)
We'll calculate the derivative of  y = arctan [x+sqrt(x^2+1)] using the chain rule
also:

y = arctan
[x+sqrt(x^2+1)]

We'll put x+sqrt(x^2+1) =
t

(arctan t)' = 1/(1+t^2)

y' =
{arctan [x+sqrt(x^2+1)]}'*[x+sqrt(x^2+1)]'

y'
= {1/{1+[x+sqrt(x^2+1)]^2}}*[1 +
2x/2sqrt(x^2+1)]

6) We'll calculate the
derivative of  y =  5arccot(t)
+ 5arccot(1/t)using the chain rule
also:

h(t)
= 5arccot(t)
+ 5arccot(1/t)

h'(t) =
-5/(1+t^2) - 5/[1+(1/t)^2]