How do you use the definition of a derivative to find the derivative of #f(x)=3x^4+2x^3-3x-2#?

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Answer 1 · 2017-02-03T09:29:15

#(df)/(dx)=12x^3+6x^2-3#

The definition of derivative gives #(df)/(dx)=Lt_(h->0)(f(x+h)-f(x))/h#

As #f(x)=3x^4+2x^3-3x-2#,

#f(x+h))=3(x+h)^4+2(x+h)^3-3(x+h)-2#

= #3(x^4+4x^3h+6x^2h^2+4xh^3+h^4)+2(x^3+3x^2h+2xh^2+h^3)-3x-3h-2#

= #3x^4+2x^3-3x-2+h(12x^3+6x^2-3)+h^2(18x^2+4x)+12xh^3+3h^4#

and #f(x+h)-f(x)=h(12x^3+6x^2-3)+h^2(18x^2+4x)+12xh^3+3h^4#

and #(df)/(dx)=Lt_(h->0)(f(x+h)-f(x))/h#

= #Lt_(h->0)1/h(h(12x^3+6x^2-3)+h^2(18x^2+4x)+12xh^3+3h^4)#

= #Lt_(h->0)12x^3+6x^2-3+h(18x^2+4x)+12xh^2+3h^3#

= #12x^3+6x^2-3#