Cracking The Ap Calculus ab Exam 2018

(remember   the shortcut    that    we  showed  you on  this    page).  To  find    the derivative  of   ,  we

first   rewrite it  as  x−3.    The derivative  of  x−3 =   −3x−4.  Therefore,  the derivative  is  

= .

18.

First,  expand  (x^2    +   8x  −   4)(2x−2 +   x−4)    to  get 2   +   16x−1   −   7x−2    +   8x−3    −   4x−4.   Now,    use the

Power   Rule    to  take    the derivative  of  each    term.   The derivative  of  2   =   0   (because    the derivative

of  a   constant    is  zero).  The derivative  of  16x−1   =   16(−1x−2)   =   −16x−2. The derivative  of  7x−2    =

7(−2x   −   3)  =   −14x−3. The derivative  of  8x−3    =   8(−3x−4)    =   24x−4.  The derivative  of  4x−4    =

4(−4x−5)     =   −16x−5.     Therefore,  the     derivative  is  −16x−2  +   14x−3   −   24x−4   +   16x−5   =  

.

19. 5 ax^4 + 4bx^3 + 3cx^2 + 2dx + e

Use the Power   Rule    to  take    the derivative  of  each    term.   The derivative  of  ax^5    =   a(5x^4 ).   The

derivative  of  bx^4    =   b(4x^3 )    =   4bx^3 . The derivative  of  cx^3    =   c(3x^2 )    =   3cx^2 . The derivative  of  dx^2

=   d(2x)   =   2   dx. The derivative  of  ex  =   e.  The derivative  of  f   =   0   (because    the derivative  of  a

constant    is  zero).  Therefore,  the derivative  is  5ax^4   +   4bx^3   +   3cx^2   +   2   dx  +   e.

20. 10(x + 1)^9

We   find    the     derivative  using   the     Chain   Rule,   which   says    that    if y    =  f(g(x)),     then   y′   =  

. Here f(x) = (x + 1)^10 . Using the Chain Rule, we get f′(x) = 10(x + 1)^9 (1) = 10(x

+   1)^9.

21.

We   find    the     derivative  using   the     Quotient    Rule,   which   says    that    if f(x)     =   ,   then   f′(x)    =  

. Here f(x) = , so u = 4x^8 − and v = 8x^4 . Using the Quotient Rule, we