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

Question

Christopher Allers · Accepted Answer

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.

Ezekiel Alexander · Answer

undefined To differentiate the expression $ \frac{x^2}{\sqrt{x+2}} - \frac{\sqrt{x+2}}{x^2} $, apply the quotient rule:

1. Differentiate the numerator and denominator separately.
2. Then apply the quotient rule formula: $ \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} $, where $ f(x) $ is the numerator and $ g(x) $ is the denominator.

Differentiating $ \frac{x^2}{\sqrt{x+2}} $:

$ f(x) = x^2 $
$ g(x) = \sqrt{x+2} $
$ f'(x) = 2x $
$ g'(x) = \frac{1}{2\sqrt{x+2}} $

Differentiating $ \frac{\sqrt{x+2}}{x^2} $:

$ f(x) = \sqrt{x+2} $
$ g(x) = x^2 $
$ f'(x) = \frac{1}{2\sqrt{x+2}} $
$ g'(x) = 2x $

Now, substitute these values into the quotient rule formula:

$ \frac{(2x)(\sqrt{x+2}) - (x^2)(\frac{1}{2\sqrt{x+2}})}{(\sqrt{x+2})^2} - \frac{(1/2\sqrt{x+2})(x^2) - (\sqrt{x+2})(2x)}{(x^2)^2} $

Simplify the expression:

$ \frac{2x\sqrt{x+2} - \frac{x^2}{2\sqrt{x+2}}}{x+2} - \frac{\frac{x^2}{2\sqrt{x+2}} - 2x\sqrt{x+2}}{x^4} $

$ \frac{4x\sqrt{x+2} - x^2\sqrt{x+2} - x^2\sqrt{x+2} + 4x\sqrt{x+2}}{2x^4} $

$ \frac{8x\sqrt{x+2} - 2x^2\sqrt{x+2}}{2x^4} $

$ \frac{4x\sqrt{x+2} - x^2\sqrt{x+2}}{x^4} $

So, the derivative of $ \frac{x^2}{\sqrt{x+2}} - \frac{\sqrt{x+2}}{x^2} $ is $ \frac{4x\sqrt{x+2} - x^2\sqrt{x+2}}{x^4} $.