What is the derivative of #sqrt(7x)#?

2 Answers
Jan 16, 2016

Answer:

#7/(2sqrt(7x)) #

Explanation:

rewrite the function as: # sqrt(7x) = (7x )^(1/2 )#

applying the 'chain rule' : # 1/2 (7x)^(-1/2 ). d/dx (7x) #

= # 1/2 (7x)^(-1/2)(7) = 7/(2sqrt(7x)) #

Jan 16, 2016

Answer:

#sqrt7/(2sqrtx)#, or, if you disapprove of radicals in the denominator, it is #sqrt(7x)/(2x)#

Explanation:

#sqrt(7x) = sqrt7sqrtx# and #d/dx(sqrtx)=1/(2sqrtx)#.

Therefore,

#d/dx(sqrt(7x)) = sqrt7d/dx(sqrtx) = sqrt7(1/(2sqrtx)) = sqrt7/(2sqrtx)#.

Eliminate the square root from the denominator if necessary.