{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW3 Solutions

# HW3 Solutions - ASSIGNMENT 3 SOLUTIONS Problem 1 We are...

This preview shows pages 1–2. Sign up to view the full content.

ASSIGNMENT 3 - SOLUTIONS Problem 1. We are told that g (2) = h (2) = 0. The product rule tells us that f (2) = g (2) h (2) + g (2) h (2) hence f (2) = 0. Problem 2. For the derivative of the first function we apply the product rule, and then the generalized power rule: ( x x 2 + 1 ) = x x 2 + 1 + x ( x 2 + 1 ) = x 2 + 1 + x 1 2 ( x 2 + 1) - 1 2 (2 x ) = x 2 + 1 + x 2 x 2 + 1 = ( x 2 + 1) + x 2 x 2 + 1 = 2 x 2 + 1 x 2 + 1 The second function is actually (2 x 2 - 1) 6 , because of the even exponent, so ( (2 x 2 - 1) 6 ) = 6(2 x 2 - 1)(2 x 2 - 1) = 24 x (2 x 2 - 1) . For the third function, the quotient rule seems like the way to go. Fortunately, we can simplify: x 16 - 1 x 4 + 1 = ( x 8 - 1)( x 8 + 1) x 4 + 1 = ( x 4 - 1)( x 4 + 1)( x 8 + 1) x 4 + 1 = ( x 4 - 1)( x 8 + 1) = x 12 - x 8 + x 4 - 1 Now x 16 - 1 x 4 + 1 = 12 x 11 - 8 x 7 + 4 x 3 . Problem 3. Consider the first function. By the generalized power rule, we have d dx ( 1 - x 2 ) = 1 2 (1 - x 2 ) - 1 2 ( - 2 x ) = - x 1 - x 2 Using the quotient rule, we obtain: d 2 dx 2 ( 1 - x 2 ) = - d dx x 1 - x 2 = - x 1 - x 2 - x ( 1 - x 2 ) 1 - x 2 We have already computed ( 1 - x 2 ) . Therefore d 2 dx 2 ( 1 - x 2 ) = - 1 - x 2 + x 2 1 - x 2 1 - x 2 = - 1 - x 2 + x 2 (1 - x 2 ) 1 - x 2 = - 1 (1 - x 2 ) 1 - x 2 . Now consider the second function. By the quotient rule, we have: d dx 2 x x - 1 = (2 x ) ( x - 1) - 2 x ( x - 1) ( x - 1) 2 = 2( x - 1) - 2 x ( x - 1) 2 = - 2 ( x - 1) 2 Hence, by the generalized power rule, we get

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}