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Let f(r)=4sq rt over r +6 (sm 3 at top sr rt symbol) sq rt over r. Then f'(r)?, f'(3)?,...

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adb2188 | (Level 3) eNoter

Posted January 31, 2013 at 3:00 AM via web

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Let f(r)=4sq rt over r +6 (sm 3 at top sr rt symbol) sq rt over r. Then f'(r)?, f'(3)?, f"(r)?, f"(30?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted January 31, 2013 at 6:14 AM (Answer #1)

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You need to find the first order derivative of the given function `f(r) = 4sqrt r + 6root(3) r` , using the chain rule, such that:

`f'(r) = 4*1/(2sqrt r) + 6*(1/3)*r^(1/3 - 1)`

`f'(r) = 2/sqrt r + 2/(root(3)(r^2))`

You need to evaluate `f'(3)` , hence, you need to substitute 3 for r such that:

`f'(r) = 2/sqrt 3 + 2/(root(3)(9))`

You need to evaluate the second order derivative, hence, you need to differentiate `f'(r)` , with respect to r,, using the quotient rule such that:

`f''(r) = (2'sqrt r - 2*(sqrt r)')/(sqrt r)^2 + (2'*(root(3)(r^2)) - 2*(root(3)(r^2))')/(r root(3)(r))`

`f''(r) = (- 2/(2sqrt r))/(sqrt r)^2 + (- 4/(3(root(3)(r))))/(r root(3)(r))`

You need to evaluate `f''(30)` such that:

`f''(30) = (-1/(sqrt 30))/30 + (- 4/(3(root(3)(30))))/(30root(3)(30))`

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