Derivative of: f(x)= (x^3+7x+5)/(1+1/x^2)

Question

Not sure How to work with the fraction in this problem. Please help! Thank you

Answer 1

Given function is f(x) = (x^3+7x+5)/(1+1/x^2)

Quioent rule in derivatives is d/dx(u/v) = (vu'-uv')/v^2

here u = x^3+7x+5  v = 1+1/x^2

d/dxf(x) = [(1+1/x^2)(3x^2+7)-(x^3+7x+5)(-2x^3)]/(1+1/x^2)^2

= [(1+1/x^2)(3x^2+7)-(-2x^6-10x^3)]/(1+1/x^2)^2

= [3x^2+7+3+7/x^2+2x^6+10x^3]/(1+1/x^2)^2

= 2x^6+10x^3+3x^2+7/x^2]//(1+1/x^2)^2

Answer 2

Given function is f(x) = (x^3+7x+5)/(1+1/x^2)

Quioent rule in derivatives is d/dx(u/v) = (vu'-uv')/v^2

here u = x^3+7x+5  v = 1+1/x^2

d/dxf(x) = [(1+1/x^2)(3x^2+7)-(x^3+7x+5)(-2x^-3)]/(1+1/x^2)^2

= [(1+1/x^2)(3x^2+7)-(-2x^6-10x^-3)]/(1+1/x^2)^2

= [3x^2+7+3+7/x^2+2x^6+10x^-3]/(1+1/x^2)^2

= 2x^6+10x^-3+3x^2+7/x^2]//(1+1/x^2)^2

= [2x^6+10/x^3+3x^2+7/x^2]/(1+1/x^4+2/x^2)