121616949-math.73 - g x = 1 p 1 √ x plugged into f x =...

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3.5 The Chain Rule 59 Now we need to compute the derivative of x x 2 + 1. This is a product, so we use the product rule: d dx x p x 2 + 1 = x d dx p x 2 + 1 + p x 2 + 1 . Finally, we use the chain rule: d dx p x 2 + 1 = d dx ( x 2 + 1) 1 / 2 = 1 2 ( x 2 + 1) - 1 / 2 (2 x ) = x x 2 + 1 . And putting it all together: f ( x ) = 2 x 2 x 2 + 1 - ( x 2 - 1)( x x 2 + 1) x 2 ( x 2 + 1) . = 2 x 2 x 2 + 1 - ( x 2 - 1) x x x 2 + 1 + x 2 + 1 x 2 ( x 2 + 1) . This can be simplified of course, but we have done all the calculus, so that only algebra is left. EXAMPLE 3.12 Compute the derivative of q 1 + p 1 + x . Here we have a more complicated chain of compositions, so we use the chain rule twice. At the outermost “layer” we have the function
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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 x-1 / 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 x-1 / 2 . = 1 8 √ x p 1 + √ x q 1 + p 1 + √ x...
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