Differentiate the function

d/d(x) [ (x

^{3}-8) • x^{2}+ 1/x^{2}- 1 )rewrite

**d/dx [ -1 + 1/x**

^{2}+ x^{2}(-8+x^{3}) ]differentiate the sum term by term

d/dx (-1) + d/dx (1/x

^{2}) + d/dx [x^{2}(-8+x^{3}) ]derivative of -1 = 0

power rule

d/dx(x

^{n}) = n•x^{n-1}n= -2, d/dx (1/x

^{2}) = d/dx (x^{-2}) = -2x^{3}**d/dx = (x**

^{2}(-8+x^{3})) + (-2/x^{3})product rule

d/dx (uv)= v du/dx + u dv/dx

u = x

^{2}, v = x^{3}-8**= (-2/x**

^{3}) + (x^{3}- 8) d/dx(x^{2}) + x^{2 }d/dx(-8+x^{3})power rule

d/dx(x

n= -2, d/dx (x

d/dx(x

^{n}) = n•x^{n-1}n= -2, d/dx (x

^{2}) = 2x**= (-2/x**

^{3}) + x^{2}(d/dx(-8+x^{3})) + (-8+x^{3})(2x)differentiate the sum term by term

**= (-2/x**

^{3}) + (2x)(-8+x^{3}) + (d/dx(-8) + d/dx(x^{3}) • x^{2}derivative of -8 is 0

Product rule

**= (-2/x**

^{3}) + (2x)(-8+x^{3}) + (3x^{2}) • x2Simplify

= (-2/x

^{3}) + (2x)(-8+x^{3}) + 3x^{4}= (-2/x

^{3}) + (-16x) + (2x^{4}) + 3x^{4}= 5 x

^{4}- 2 / x^{3}- 16 x
## Comments

^{2}+ x^{2}(-8+x^{3}) ] is what she meant?