MATH 41 Midterm: Stanford MATH 41 exam2sol

• School: Stanford University
• Department: MATH - Mathematics
• Course: MATH 41
• Professor: All

Document Summary

So by chain rule du dx df (x) dx df (u) du du dx dx(cid:19) =(cid:0)2013 cos2012 u ( sin u)(cid:1)(cid:18) du. = 2013 cos2012(x arctan x + 4 ) sin(x arctan x + 4 ) (cid:18)arctan x + x x2 + 1(cid:19) (b) g(t) = log3(cid:0)sec(cid:0)10 t(cid:1)(cid:1) (5 points) by the base-change formula, g(t) = we get ln(sec(cid:0)10 t(cid:1)) ln 3. 1 d dt sec(cid:0)10 t(cid:1) sec(cid:0)10 t(cid:1) tan(cid:0)10 t(cid:1) sec(cid:0)10 t(cid:1) tan(cid:0)10 t(cid:1) 10 t ln 10. 1 ln 3 sec (10 t) ln 3 sec (10 t) ln 3 sec (10 t) T (c) h(z) = zln z + arcsin(cid:0)z2 1(cid:1) (5 points) let f (z) = zln z and g(z) = arcsin(z2 1). Then using logarithmic di erentiation ln f (z) = ln(zln z) = (ln z)(ln z) = (ln z)2 and using chain rule g (z) =