How do you differentiate #x^2/(sqrt(x+2))-sqrt(x+2)/x^2#?

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Answer 1 · 2017-08-13T21:16:17

Please see below.

Let #y = x^2/(sqrt(x+2))-sqrt(x+2)/x^2#

Note that with #u = x^2/sqrt(x+2)#, we have

#y = u-1/u#.

So #y' = u'+1/u^2 u' = u'(1+1/u^2)#

Use the quotient rule to find

#u' = ((2x)sqrt(x+2)-x^2(1/(2sqrt(x+2))))/(sqrt(x+2))^2#

# = (4x(x+2)-x^2)/(2(x+2)^(3/2))#

# = (3x^2+8x)/(2(x+2)^(3/2))#

So

#y' = (3x^2+8x)/(2(x+2)^(3/2))(1+1/(x^2/sqrt(x+2))^2)#

# = (3x^2+8x)/(2(x+2)^(3/2))(1+(x+2)/x^4)#

Simplify further to taste.