What is the derivative of #1/sqrtx#?

1 Answer
Jun 2, 2015

Oe law of exponentials states that #a^-n=1/a^n#. Also, another law states that #a^(m/n)=root(n)(a^m)#. Finally, we'll need a third one: #(a^n)^m=a^(n*m)#

Thus, we can rewrite this function following both laws. Then, we derivate it.

First law applied:

#(sqrtx)^-1#

Second law:

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

Third law mentioned:

#x^(-1/2)#

Now, deriving #f(x)=x^(-1/2)#

#(dy)/(dx)=(-1/2)x^(-3/2)=-1/(2x^(3/2))#

Or, if you prefer the root: #-1/(2sqrt(x^3))#