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

Click to edit the document details