What is the derivative of #(x-1)/sqrtx#?

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Answer 1 · 2015-07-11T10:28:27

#d/dx((x-1)/sqrt(x))=(x+1)/(2xsqrt(x))#

Because the square root function comes up a lot (Pyhthagorean Theorem, diagonal distance), it is helpful to memorize (eventually):

#d/dx(sqrtx) = 1/(2sqrtx)#

For #y = (x-1)/sqrtx#, using the quotient rule, we get:

#d/dx((x-1)/sqrt(x))=(sqrtx-(x-1)(1/(2sqrtx)))/x=(sqrtx-x/(2sqrtx)+1/(2sqrtx))/x#

#=(sqrtx-sqrtx/2+1/(2sqrtx))/x = ((sqrtx - sqrtx/2+1/(2sqrtx))sqrtx)/(xsqrtx)#

#=(x-x/2+1/2)/(xsqrtx)= ((x/2+1/2)2)/((xsqrtx)2)#

#(x+1)/(2xsqrtx)#

Answer 2 · 2015-07-11T12:28:52

The quotient rule will work, but , if we prefer, we can rewrite the expression before differentiating.

#y = (x-1)/x^(1/2) = x/x^(1/2) -1/x^(1/2)#

# = x^(1/2) - x^(-1/2)#

Now we use the power rule:

#y' = 1/2(x^(-1/2)) -(-1/2)x^(-3/2)#

# = 1/2x^(-1/2) + 1/2x^(-3/2)#

Which may be rewritten using ratios or as a single ratio:

#y' = 1/(2x^(1/2)) +1/(2x^(3/2))#

# = x/(2x^(3/2)) +1/(2x^(3/2)) = (x+1)/(2x^(3/2))#

If we would rather have radicals than rational exponents, we can write:

#y' = (x+1)/(2xsqrtx)#