How do you find the derivative of #sqrt(x-3)/sqrt(x+3)#?

1 Answer
Nallasivam V
Apr 2, 2018

#dy/dx=[6]/(2sqrt(x-3)(x+3)^(3/2))#

Explanation:

Given -

#y=sqrt(x-3)/sqrt(x+3)#

#dy/dx={[sqrt(x+3)(1/(2sqrt(x-3)))] -[sqrt(x-3)(1/(2sqrt(x+3)))]}/(sqrt(x+3))^2#

#dy/dx={[sqrt(x+3)/(2sqrt(x-3))]-[(sqrt(x-3))/(2sqrt(x+3))]}/(x+3)#

#dy/dx={sqrt(x+3)/(2sqrt(x-3))-(sqrt(x-3))/(2sqrt(x+3))}/(x+3)#

#dy/dx={[(sqrt(x+3)sqrt(x+3))-(sqrt(x-3)sqrt(x-3))]/(2sqrt(x-3)sqrt(x+3))}/(x+3)#

#dy/dx={[(x+3)-(x-3)]/(2sqrt(x-3)sqrt(x+3))}/(x+3)#

#dy/dx={[x+3-x+3]/(2sqrt(x-3)sqrt(x+3))}/(x+3)#

#dy/dx={[cancelx+3cancel(-x)+3]/(2sqrt(x-3)sqrt(x+3))}/(x+3)#

#dy/dx={[6]/(2sqrt(x-3)sqrt(x+3))}/(x+3)#

#dy/dx={[6]/(2sqrt(x-3)sqrt(x+3))}-:(x+3)#

#dy/dx={[6]/(2sqrt(x-3)sqrt(x+3)(x+3))}#

#dy/dx=[6]/(2sqrt(x-3)(x+3)^(3/2))#

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