Cracking The Ap Calculus ab Exam 2018

(Marvins-Underground-K-12) #1
Now,    use the Power   Rule    to  take    the derivative  of  each    term.   The derivative  of  x   =    x .    The

derivative  of  x   =    x .    Therefore,  the derivative  is      =   .

SOLUTIONS TO PRACTICE PROBLEM SET 5


1.

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

. Here, f(x) = , so u = 4x^3 − 3x^2 and v = 5x^7 +. Using the Quotient Rule,


we  get f′(x)   =    .  This    can be  simplified  to  f′

(x) =   .


  1. 3 x^2 − 6x − 1


We  find    the derivative  using   the Product Rule,   which   says    that    if  f(x)    =   uv, then    f′(x)   =   u   +   v

. Here f(x) = (x^2 − 4x + 3)(x + 1), so u = x^2 − 4x + 3 and v = x + 1. Using the Product Rule,


we  get f′(x)   =   (x^2    −   4x  +   3)(1)   +   (x  +   1)(2x   −   4). This    can be  simplified  to  f′(x)   =   3x^2    −   6x  −   1.

3.

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

. Here f(x) = 8 , which can be written as f(x) = 8(x^4 − 4x^2 ). Using


the Chain   Rule,   we  get f′(x)   =   8 (x^4  −   4x^2 ) (4x^3    −   8x).    This    can be  simplified  to  f′(x)   =   

.

4.
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