What is the derivative of #5/sqrtx#?

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Answer 1 · 2017-01-12T18:01:47

I got: #f'(x)=-5/(2xsqrt(x))#

We can use the fact that #1/sqrt(x)=x^(-1/2)# and derive as usual as an exponent:
#f'(x)=5*-1/2*x^(-1/2-1)=-5/2x^(-3/2)=-5/(2x^(3/2))=-5/(2xsqrt(x))#

Answer 2 · 2017-01-12T18:04:41

#(-5)/2 (x^(-3/2))#

#d/dx 5/sqrtx =d/dx 5 x^(-1/2)#

=#5 (-1/2)(x^(-3/2))#

#(-5)/2 (x^(-3/2))#