What is the derivative of 3x² using first principle?

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Answer 1 · 2018-01-16T21:03:30

See explanation.

According to the definition of derrivative a limit:

#lim_{h->0}(f(x_0+h)-f(x_0))/h# (if exists) is called the derrivative #f'(x_0)# of the function #f(x)#

In the case given we have:

#lim_{h->0}(3(x_0+h)^2-3x_0^2)/h=#

#=lim_{h->0}(3*(x_0^2+2x_0h+h^2)-3x_0^2)/h=#

#=lim_{h->0}(cancel(3x_0^2)+6x_0h+3h^2 cancel(-3x_0^2))/h=#

#=lim_{h->0} (6x_0h+3h^2)/h=lim_{h->0} (cancel(h)(6x_0+3h))/cancel(h)=#

#=lim_{h->0}(6x_0+3h)=6x_0#

So we used the First Principle to calculate that the derivative of #f(x)=3x^2# is #f'(x)=6x#