How do you find the derivative of #2/sqrtx#?

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Answer 1 · 2015-06-03T21:02:28

One law of exponentials states that #a^(m/n)=root(n)(a^m)#. Thus, we can rewrite the function as

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

Also, another law states that #a^-n=1/a^n#

Thus, we can rewrite this again as #2x^(-1/2)#

Now, derivating the function:

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

If you prefer, #-1/sqrt(x^3)#