What is the derivative of #sqrt(2x)# ?

2 Answers
George C.
Mar 26, 2017

#d/(dx) sqrt(2x) = sqrt(2)/(2sqrt(x))#

Explanation:

For any non-zero power #p# of #x# we have:

#d/(dx) x^p = p x^(p-1)#

So:

#d/(dx) sqrt(2x) = d/(dx) sqrt(2)sqrt(x)#

#color(white)(d/(dx) sqrt(2x)) = d/(dx) sqrt(2)x^(1/2)#

#color(white)(d/(dx) sqrt(2x)) = sqrt(2)*1/2x^(-1/2)#

#color(white)(d/(dx) sqrt(2x)) = sqrt(2)/2*1/x^(1/2)#

#color(white)(d/(dx) sqrt(2x)) = sqrt(2)/2*1/sqrt(x)#

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

George C.
Mar 26, 2017

#d/(dx) sqrt(2x) = sqrt(2)/(2sqrt(x))#

Explanation:

Here is how to find it using the limit definition of derivative...

#d/(dx) sqrt(2x) = lim_(h->0) (sqrt(2(x+h))-sqrt(2x))/h#

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

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

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

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

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

#color(white)(d/(dx) sqrt(2x)) = 2/(sqrt(2(x+0))+sqrt(2x))#

#color(white)(d/(dx) sqrt(2x)) = 1/(sqrt(2x))#

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

Related questions
See all questions in Chain Rule
Impact of this question
6414 views around the world
You can reuse this answer
Creative Commons License