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How do you find the derivative of #g(x)=1/sqrtx#?

1 Answer

Aug 23, 2017

see below

Explanation:

#g(x)=1/sqrtx#

#=>d/dxg(x)=d/dx(x)^(-1/2)#

Utilize the formula,

#d/dxx^n=nx^(n-1)#

#=>d/dx(x)^(-1/2)=-1/2x^((-1/2-1))=-1/(2xsqrtx)#

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