`f(x) = 1/(sqrt(x + 4)), (0,(1/2))` Evaluate the second derivative of the function at the given point. Use a computer algebra system to verify your result.

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Chapter 2, 2.4 - Problem 92 - Calculus of a Single Variable (10th Edition, Ron Larson).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the second derivative of the function, hence, you need first to evaluate the first derivative, using the quotient and the chain rules, such that:

`f'(x) = (1'*sqrt(x+4) - 1*(sqrt(x+4))')/((sqrt(x+4))^2)`

`f'(x) = (0*sqrt(x+4) - 1/(2sqrt(x+4)))/(x+4)`

`f'(x) = (- 1/(2sqrt(x+4)))/(x+4)`

`f'(x) = -1/(2(x+4)sqrt(x+4))`

`f'(x) = -1/(2sqrt(x+4)^3)`

`f'(x) = -1/(2(x+4)^(3/2))`

You may evaluate the second derivative of the function using the quotiemt and the chain rules:

`f''(x) = (-1'*(2(x+4)^(3/2)) +1*(2(x+4)^(3/2))')/(4(x+4)^3)`

`f''(x) = (0+2*(3/2)(x+4)^(3/2-1))/(4(x+4)^3)`

`f''(x) = (3sqrt(x+4))/(4(x+4)^3)`

You need to evaluate the second derivative at the point (0,1/2), hence, you need to replace 0 for x in equation `f''(x) = (3sqrt(x+4))/(4(x+4)^3)` :

`f''(0) = (3sqrt(0+4))/(4(0+4)^3)`

`f''(0) = (3sqrt(4))/(4*(4)^3) => f''(0) = (6)/(256) => f''(0) = 3/128`

Hence, evaluating the second derivative at (0,1/2), yields `f''(0) = 3/128.`

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