Cracking The Ap Calculus ab Exam 2018

    =    .  This    simplifies  to  f′(x)   =       =    .  Now

we  can cancel  the factor  h   in  the numerator   and the denominator to  get f′(x)   =       =   

. Now we take the limit: f′(x) = .
16. 2 ax + b

We  find    the derivative  of  a   function,   f(x),   using   the definition  of  the derivative, which   is  f′(x)   =   

. Here f(x) = ax^2 + bx + c and f(x + h) = a(x + h)^2 + b(x + h) + c = ax^2 +

2 axh   +   ah^2    +   bx  +   bh  +   c.

If  we  now plug    these   into    the definition  of  the derivative, we  get f′(x)   =   

=    .   This    simplifies  to  f′(x)   =  

. Now we can factor out the h from the numerator and cancel it with the h

in  the denominator:    f′(x)   =       =    (2ax   +   ah  +   b).

Now we  take    the limit   to  get f′(x)   =       (2ax    +   ah  +   b)  =   2ax +   b.

SOLUTIONS TO PRACTICE PROBLEM SET 4

1. 64 x^3 + 16x

First,  expand  (4x^2   +   1)^2    to  get 16x^4   +   8x^2    +   1.  Now,    use the Power   Rule    to  take    the derivative

of  each    term.   The derivative  of  16x^4   =   16(4x^3 )   =   64x^3 . The derivative  of  8x^2    =   8(2x)   =   16x.

The derivative  of  1   =   0   (because    the derivative  of  a   constant    is  zero).  Therefore,  the derivative

is  64x^3   +   16x.

2. 10 x^9 + 36x^5 + 18x

First,  expand  (x^5    +   3x)^2   to  get x^10    +   6x^6    +   9x^2 .  Now,    use the Power   Rule    to  take    the derivative

of   each    term.   The     derivative  of x^10 =   10x^9 .     The     derivative  of  6x^6 =  6(6x^5 )    =   36x^5 .     The

derivative  of  9x^2    =   9(2x)   =   18x.    Therefore,  the derivative  is  10x^9   +   36x^5   +   18x.

3. 77 x^6