How do you find f'(x) using the definition of a derivative for #f(x)=sqrt(1+2x)#?

1 Answer
George C.
Oct 8, 2015

Use definition:

#f'(a) = lim_(h->0) (f(a+h) - f(a))/h#

to find: #f'(x) = 1/sqrt(1+2x)#

Explanation:

Let #f(x) = sqrt(1+2x)#

Then the derivative at #x=a# is defined as the following limit:

#f'(a) = lim_(h->0) (f(a+h) - f(a))/h#

#= lim_(h->0) (sqrt(1+2(a+h)) - sqrt(1+2a))/h#

#= lim_(h->0) (sqrt(1+2(a+h)) - sqrt(1+2a))/h * (sqrt(1+2(a+h)) + sqrt(1+2a))/(sqrt(1+2(a+h)) + sqrt(1+2a))#

#= lim_(h->0) (sqrt(1+2(a+h)) - sqrt(1+2a))/h * (sqrt(1+2(a+h)) + sqrt(1+2a))/(sqrt(1+2(a+h)) + sqrt(1+2a))#

#= lim_(h->0) ((1+2(a+h)) - (1+2a))/(h(sqrt(1+2(a+h)) + sqrt(1+2a)))#

#= lim_(h->0) (2color(red)(cancel(color(black)(h))))/(color(red)(cancel(color(black)(h)))(sqrt(1+2(a+h)) + sqrt(1+2a)))#

#= lim_(h->0) 2/(sqrt(1+2(a+h)) + sqrt(1+2a))#

#= 2/(sqrt(1+2a) + sqrt(1+2a))#

#= 1/sqrt(1+2a)#

So #f'(x) = 1/sqrt(1+2x)#

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