How do you find the derivative of #f(x)=ax^2#?

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Answer 1 · 2017-01-09T15:45:27

#d/(dx) (ax^2) = 2ax#

This is an application of a basic differentiation rule:

#d/(dx) x^n = n*x^(n-1)#

In this case:

#d/(dx) (ax^2) = a* d/(dx) (x^2) = 2ax#

You can also demonstrate it directly considering the formal definition of the derivative of #f(x)#:

#f'(x) = lim_(Deltax->0) (f(x+Deltax)-f(x))/(Deltax) = lim_(Deltax->0) (Deltaf)/(Deltax)#

#Delta f = a(x+Deltax)^2 - ax^2 = ax^2+2axDeltax +a(Deltax)^2-ax^2=2axDeltax +a(Deltax)^2#

#(Deltaf)/(Deltax) = 2ax +aDeltax#

#lim_(Deltax->0) (Deltaf)/(Deltax) = lim_(Deltax->0)2ax +aDeltax = 2ax#