Using the limit definition, how do you differentiate #f(x)=x^3 + x#?

1 Answer
Nov 14, 2015

#dy/dx = 3x^2 + 1#

Explanation:

We have

#y = x^3 + x#

So

#dy/dx = lim_(Deltax rarr 0)(Deltay)/(Deltax)#

For ease of typing let's say #Deltax = h#

#dy/dx = lim_(h rarr 0)((x+h)^3 + x + h - x^3 - x)/h#

#dy/dx = lim_(h rarr 0)(x^3 + 3x^2h + 3xh^2 + h^3 + x + h - x^3 - x)/h#

#dy/dx = lim_(h rarr 0)(cancel(x^3) + 3x^2h + 3xh^2 + h^3 cancel(+x) + h cancel(- x^3) cancel(- x))/h#

#dy/dx = lim_(h rarr 0)(3x^2h + 3xh^2 + h^3 + h)/h#

#dy/dx = lim_(h rarr 0)3x^2 + 3xh + h^2 + 1#

#dy/dx = 3x^2 + 1#