Question #79c72

Question

1259 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-08T23:05:16

#2x^-3# or #2/x^3#

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

power rule: #y' = nx^(n-1)#, where #n = -1#

#y' = -1 * x^-2 = -1x^-2# or #-1/x^2#

#y'' = (deltay)/(deltax) (-x^-2)#

#y'' = nx^(n-1)#, where #n = -2#
#y'' = -2 * -x^(-3) = 2x^-3# or #2/x^3#

Answer 2 · 2018-01-08T23:05:25

Differentiate twice.
Answer #2x^-3#

First we need to simplify the expression we are truing to differentiate.

From Laws of indices, we know that #1/x = x^-1#

Now to differentiate, we multiply the coefficient by the power, and then #-1# from the power.

#dy/dx = nax^n-1# where a## is the coefficient and #n# is the power.

#therefore dy/dx x^-1 = -1x^-2 = -x^-2#

However, we are finding the second derivative so in turn we must differentiate once more.

#therefore (d^2y)/dx^2 = 2x^-3#