Unformatted text preview: g ( x ) = 1 + p 1 + √ x plugged into f ( x ) = √ x , so applying the chain rule once gives d dx r 1 + q 1 + √ x = 1 2 ± 1 + q 1 + √ x ¶1 / 2 d dx ± 1 + q 1 + √ x ¶ . Now we need the derivative of p 1 + √ x . Using the chain rule again: d dx q 1 + √ x = 1 2 ( 1 + √ x )1 / 2 1 2 x1 / 2 . So the original derivative is d dx r 1 + q 1 + √ x = 1 2 ± 1 + q 1 + √ x ¶1 / 2 1 2 ( 1 + √ x )1 / 2 1 2 x1 / 2 . = 1 8 √ x p 1 + √ x q 1 + p 1 + √ x...
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 Spring '07
 JonathanRogawski
 Math, Calculus, Chain Rule, Derivative, Product Rule, The Chain Rule, dx

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