How do you use the limit definition to find the derivative of #f(x)=2x^2+1#?

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How do you use the limit definition to find the derivative of #f(x)=2x^2+1#?

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Answer 1 · 2016-11-02T21:36:15

# f'(x) = 4x#

Explanation:

By definition, # f'(x)= lim_(h->0)(f(x+h)-f(x))/h#

so for # f(x) = 2x^2 +1 # we have:
# f'(x) = lim_(h->0)( { (2(x+h)^2 +1) - (2x^2 +1)} )/h #
# f'(x) = lim_(h->0)( (2(x^2+2hx+h^2) +1 ) - 2x^2+1)/h#
# f'(x) = lim_(h->0)( 2x^2+4hx+2h^2 +1 - 2x^2-1 )/h#
# f'(x) = lim_(h->0)( 4hx+2h^2 )/h#
# f'(x) = lim_(h->0)( 4x+2h)#
# f'(x) = 4x#

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How do you use the limit definition to find the derivative of #f(x)=2x^2+1#?

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