Derivative with respect to 'x' of (5x)1/2 = (1/2) (5x)-1/2 (5) = 2.5/sqrt(5x)
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a+ square root of b has a conjugate a- square root of b and this is used rationalize the denominator when it contains a square root. If we want to multiply 5 x square root of 10 by something to get rid of the radical you can multiply it by square root of 10. But if we look at 5x( square root of 10 as ) 0+ 5x square root of 10 then the conjugate would be -5x square root of 10
5x
f'(x)=.5x^-.5 f''(x)=-.25x^-1.5 f'''(x)=.375x^-2.5 or f'''(x)=(3/8)x^(-5/2)
Call this number "x". In this case:x = root(5 root(5 root(5 root(5... Since the expression inside the first root is equal to the entire expression, you get: x = root(5 x) Squaring both sides, to get rid of the rood, you get: x squared = 5x x squared - 5x = 0 x(x - 5) = 0 So, x is either equal to 0, or to 5. Indeed, if you start with any number and repeatedly multiply by 5 and then take the square root, you get closer and closer to 5... unless you start with zero, in which case you get exactly zero.
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