How do you differentiate #f(x)= (x^3+2x+1)(2+x^-2)#?

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Answer 1 · 2015-05-19T14:56:56

This is a simple and delicious case of chain rule, mate!

Let #u=x^3+2x+1# and #v=2+x^-2# and remember that the chain rule states that

When #y=f(x)g(x)#, then #y'=f'(x)g(x)+f(x)g'(x)#, Let's proceed this way.

#y'=u'v+v'u#

But #u'=3x^2+2# and #v'=-2x^-3#

So,

#y'=(3x^2+2)(2+x^-2)+(-2x^-3)(x^3+2x+1)#

Let's just distribute these products in order to try to make it simpler:

#y'=(6x^2+3+4+2x^-2)+(-2-4x^-2-2x^-3)#

Reorganizing,

#y'=6x^2-2x^-2-2x^-3+5#