Cracking The Ap Calculus ab Exam 2018

(2  csc x^2 )(−csc  x^2     cot x^2 )(2x)   =   −4x csc^2 (x^2 )cot(x^2 ).

7. 16 sin 2x

Recall  that     (sin   x)  =   cos x   and that     (cos   x)  =   −sin    x.  Here    we  will    use the Chain   Rule    four

times   to  find    the fourth  derivative.

The first   derivative  is      =   (cos    2x)(2)  =   2   cos 2x.

The second  derivative  is      =   2(−sin  2x)(2)  =   −4  sin 2x.

The third   derivative  is      =   −4(cos  2x)(2)  =   −8  cos 2x.

And the fourth  derivative  is      =   −8(−sin 2x)(2)  =   16  sin 2x.

8. cos (1 = cos^2 x) + sin(1 + cos^2 x)

Recall  that     (sin   x)  =   cos x   and that     (cos   x)  =   −sin    x.  Here    we  will    use the Chain   Rule    to

find    the derivative:     =   cos t   −   (−sin   t)  =   cos t   +   sin t   and     =   2(cos   x)(−sin x)  =   −2  sin x

cos x.  Next,   because     and t   =   1   +   cos^2 x,    we  get     =   (cos    t   +   sin t)(−2   sin x   cos

x)  =   [cos(1  +   cos^2 x)    +   sin(1   +   cos^2 x)](−2    sin x   cos x).

9.

Recall  that        (tan    x)  =   sec^2 x.    Using   the Quotient    Rule    and the Chain   Rule,   we  get     =   

. This simplifies to .
10. −[sin(1 + sin θ)](cos θ)

Recall  that     (sin   x)  =   cos x   and that     (cos   x)  =   −sin    x.  Using   the Chain   Rule,   we  get     =   −

[sin(1  +   sinθ)](cosθ).