How do you find the third derivative of #y=x^2-2/x#?

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Answer 1 · 2018-04-05T20:39:44

#=>f'''(x)=12/x^(4)#

We have:

#f(x)=x^2-2/x# We can rewrite this as:

#f(x)=x^2-2x^-1# We use the power rule:

#d/dx[x^n]=nx^(n-1)# if #n# is a constant.

#f'(x)=d/dx[x^2]-d/dx[2x^-1]#

#=>f'(x)=d/dx[x^2]-2*d/dx[x^-1]#

#=>f'(x)=2*x^(2-1)-2*-1*x^(-1-1)#

#=>f'(x)=2x-2*-x^(-2)#

#=>f'(x)=2x+2x^(-2)# Repeat the process.

#=>f''(x)=d/dx[2x]+d/dx[2x^(-2)]#

#=>f''(x)=2*1*x^(1-1)+2*-2*x^-3#

#=>f''(x)=2-4x^-3# Again!

#=>f'''(x)=d/dx[2]-d/dx[4x^-3]#

#=>f'''(x)=0-4*-3*x^(-3-1)#

#=>f'''(x)=0+12*x^(-4)#

#=>f'''(x)=12*1/x^(4)#

#=>f'''(x)=12/x^(4)#