# Find the derivative of the function. Simplify if possible. and...... y = 17 arctan(sqrt x) y = arcsin(4x + 2) y = arccos(e8x) h(t) = 5arccot(t) + 5arccot(1/t)

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### 2 Answers

To find the derivative of

1)y = 17 arctan sqrtx

tan (y/17) = sqrtx

Differentiating with respect to x, weget:

sec^2 (y/17) * (1/17) dy/dx = (1/2sqrtx)

dy/ dx = (17/2sqrtx){1+(tan(y/17))^2}

dy/dx = (17/2sqrtx) {1+x}

2)y = arc sin(4x+2)

Therefore siny = 4x+2

Differentiate both sides:

cosy*dy/dx = d/dx(4x+2)

cosy * dy/dx = 4

dy/dx = 4/cosy = 4/sin^2y = 4/(4x+2)^2

dy/dx = 4/(4x+2)^2.

3)y=arccos(e^8x).

Therefore cosy = e^(8x).

Differentiating both sodes, we get:

-siny*dy/dx= d/dx{e^(8x)}.

-siny*dy/dx = e^(8x)* d/dx(8x).

-siny*dy/dx = 8e^(8x).

dy/dx = 8e^(8x)* (1/-siny)

dy/dx = -8e^(8x)/ (1-(cosy)^2)

dy/dx = -8e^(8x)/{1- e^(16x)}, as (cosy)^2 = (e^(8x))^2.

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(4*x* + 2)using the chain rule also:

y = [arcsin(4*x* + 2)]'*(4x+2)'

We'll see 4*x* + 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^*8*x*) using the chain rule also:

*y* = arccos(*e^*8*x*)

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

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

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

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