Solved PYQs 515-523 of 2012: AP (Advanced Placement) Calculus AB

${\displaystyle -{\displaystyle \left(\frac{1}{2}\right)}}$

${\displaystyle -2}$

${\displaystyle -\log 2}$

${\displaystyle -\left[\frac{1}{{\displaystyle 2\log 2}}\right]}$

${\displaystyle x{e}^x}$

${\displaystyle -x{e}^{{\displaystyle -x}}}$

${\displaystyle -x{e}^x}$

${\displaystyle x{e}^{{\displaystyle -x}}}$

${\displaystyle -x\cos {\displaystyle \left(\log x\right)}}$

${\displaystyle \sin {\displaystyle \left(\log x\right)}-\cos {\displaystyle \left(\log x\right)}}$

${\displaystyle \sin {\displaystyle \left(\log x\right)}+\cos {\displaystyle \left(\log x\right)}}$

${\displaystyle +x\sin {\displaystyle \left(\log x\right)}}$

${\displaystyle \left[\frac{{{\displaystyle (x+3)}}^9}{9}\right]-\left[\frac{{{\displaystyle (x+3)}}^8}{8}\right]}$

${\displaystyle \left[\frac{{\displaystyle \left\{{{\displaystyle (x+3)}}^7{\displaystyle (8x+33)}\right\}}}{72}\right]}$

${\displaystyle \left[\frac{{\displaystyle \left\{{{\displaystyle (x+3)}}^8{\displaystyle (8x+33)}\right\}}}{72}\right]}$

${\displaystyle \left[\frac{{{\displaystyle (x+3)}}^8}{8}\right]}$

${\displaystyle 2{\tan }^{{\displaystyle -1}}\sqrt{{\displaystyle (x+2)}}}$

${\displaystyle 2{\tan }^{{\displaystyle -1}}\sqrt{{\displaystyle (x^2+2)}}}$

${\displaystyle 2{\tan }^{{\displaystyle -1}}x}$

${\displaystyle 2{\tan }^{{\displaystyle -1}}\sqrt{{\displaystyle (x^2+3)}}}$

${\displaystyle \left[\frac{e^x}{{{\displaystyle (x+1)}}^2}\right]}$

${\displaystyle \left[\frac{{\displaystyle (-e^x)}}{{{\displaystyle (x+1)}}^2}\right]}$

${\displaystyle \left[\frac{e^x}{{\displaystyle (x+1)}}\right]}$

${\displaystyle \left[\frac{{\displaystyle x{e}^x}}{{{\displaystyle (x+1)}}^2}\right]}$

${\displaystyle -\log \left[\frac{{\displaystyle e^x+1}}{e^x}\right]}$

${\displaystyle \log \left[\frac{{\displaystyle e^x+1}}{{\displaystyle 2e^x}}\right]}$

${\displaystyle -\log \left[\frac{e^x}{{\displaystyle e^x+1}}\right]}$

${\displaystyle \log \left[\frac{{\displaystyle 2e^x}}{{\displaystyle e^x+1}}\right]}$

${\displaystyle \left[\frac{{\displaystyle \sin 2x}}{2}\right]}$

${\displaystyle -\left[\frac{{\displaystyle \cos 2x}}{2}\right]}$

${\displaystyle \left[\frac{{\displaystyle \cos 2x}}{2}\right]}$

${\displaystyle -\left[\frac{{\displaystyle \sin 2x}}{2}\right]}$

${\displaystyle {\tan }^{{\displaystyle -1}}{\displaystyle \left(\log x\right)}}$

${\displaystyle \left[\frac{{\displaystyle {\tan }^{{\displaystyle -1}}x}}{x}\right]}$

${\displaystyle \left[\frac{{\displaystyle \left\{{\tan }^{{\displaystyle -1}}{\displaystyle \left(\log x\right)}\right\}}}{x}\right]}$

${\displaystyle {\tan }^{{\displaystyle -1}}x}$