What is the second derivative of #x^2 + (16/x)#?

1 Answer
Steve M
Oct 29, 2016

# :. d^2/dx^2( x^2 + 16/x) = 2 +32/x^3 #

Explanation:

You should learn the power rule for differentiation, which is that:
# d/dx(x^n) = nx^(n-1) AA n in RR #

So, # d/dx( x^2 + 16/x) = d/dx (x^2 + 16x^-1)#
# :. d/dx( x^2 + 16/x) = 2x^(2-1) + 16(-1)x^(-1-1)#
# :. d/dx( x^2 + 16/x) = 2x -16x^-2#

And differentiating a second time, gives us:
# d^2/dx^2( x^2 + 16/x) = 2(1)x^(1-0) -16(-2)x^(-2-1)#
# :. d^2/dx^2( x^2 + 16/x) = 2x^0 +32x^-3 #
# :. d^2/dx^2( x^2 + 16/x) = 2 +32/x^3 #

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