Cracking The Ap Calculus ab Exam 2018

. Here = and = 3x^2 . Now we plug x = 1 into the

derivative. Note    that    where   x   =   1,  t   =   (1)^3   =   1.  We  get     =       =   −8  and 

    =   3(1)^2  =   3.  Thus,       =   (−8)(3) =   −24.

18. +

Here    the solution    will    be  much    simpler if  we  first   substitute  x   =       into    the expression  for y.

We  get y   =    .  Now we  find    the derivative  using   the Product Rule,   which   says

that    if  f(x)    =   uv, then    f′(x)   =   u   +   v . Here    y   =    ,  so  u   =   t^3     −6t and v   =   5t  +   

. Using the Product Rule, we get = + .

19. 2

We  find    the derivative  using   the Chain   Rule,   which   says    that    if  y   =   y(v)    and v   =   v(x),   then    

. Here and = 2x. Now, we plug x = 1 into the

derivative. Note    that,   where   x   =   1,  u   =   0.  We  get     =   1   and     =   2,  so  

= (1)(2) = 2.

20.

We  find    the derivative  using   the Chain   Rule,   which   says    that    if  y   =   y(v)    and v   =   v(x),   then    

,   although,   in  this    case,   we  have    u(y),   y(x),   and x(v),   so  we  will    find        by  

. Here = 3y^2 , = , and = 2v. Next,

    =    .  Now because x   =   v^2     and y   =    ,  we  get     =

(2v)    =   .