What is the derivative of #y'# given #y = xsqrt(x) + 1/(x^2 sqrt(x))# ?

1 Answer
Jul 12, 2016

#y'' = 1/4(3x^(-1/2) +35 x^(-9/2))#

Explanation:

#y = xsqrt(x) + 1/(x^2 sqrt(x))#

We are asked to find the derivative of #y'# which is the second derivative of #y#, represented by #(d^2y)/dx^2# or #y''#

First rewrite the expression for #y# in term of powers of x #->#

#y = x^(3/2) + x^(-5/2)#

Using the power rule #->#

#y' = 3/2x^(1/2) - 5/2 x^(-7/2)#

Using the power rule again #->#

#y'' = 3/2 * 1/2 x^(-1/2) - 5/2 * (-7/2) x^(-9/2)#

#y'' = 3/4 x^(-1/2) + 35/4 x^(-9/2)#

#y'' = 1/4(3x^(-1/2) +35 x^(-9/2))#