What is the derivative of #x^2# ?

2 Answers
smendyka
Jul 27, 2017

See a solution process below:

Explanation:

For #x^color(red)(2)#:

The directive of #x^color(red)(a)# is #color(red)(a)x^(color(red)(a)- 1)#

For #x^color(red)(2)# the derivative is:

#d/dx x^color(red)(2) = color(red)(2)x^(color(red)(2)- 1) = 2x^1 = 2x#

George C.
Jul 27, 2017

#d/dx x^2 = 2x#

Explanation:

The derivative of a function is essentially its slope at a point.

If you consider the slope of a secant between a point on the graph of the function and a nearby point, then consider the limit as you bring that second point closer and closer to the original point, the limit is the slope of the tangent to the graph of the function at that point.

So with #f(x) = x^2# we find:

#d/(dx) f(x) = lim_(h->0) (f(x+h)-f(x))/((x+h)-x)#

#color(white)(d/(dx) f(x)) = lim_(h->0) (f(x+h)-f(x))/h#

#color(white)(d/(dx) f(x)) = lim_(h->0) ((x+h)^2-x^2)/h#

#color(white)(d/(dx) f(x)) = lim_(h->0) (color(red)(cancel(color(black)(x^2)))+2hx+h^2-color(red)(cancel(color(black)(x^2))))/h#

#color(white)(d/(dx) f(x)) = lim_(h->0) (h(2x+h))/h#

#color(white)(d/(dx) f(x)) = lim_(h->0) (2x+h)#

#color(white)(d/(dx) f(x)) = 2x#

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