MATH 41 Midterm: Stanford MATH 41 10exam2sol

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

Document Summary

Now, we do each separate derivative using the chain rule: d dt ln(sec t + tan t) = and d dt log2(2 + t) = d dt ln(2 + t) ln 2. 1 d dt sec t + tan t sec t tan t + sec2 t sec t + tan t (sec t + tan t) Combining, we get that p (t) = sec t + 1 (ln 2)(2 + t) (b) f (x) = 10sin( x) +qx + x (4 points) again, we decompose using the rule for a sum: f (x) = d dx. For the rst term, using the chain rule and the fact that d dx (ax) = ln a ax, we get d dx. 10sin( x) = ln 10 10sin( x) d sin( x) = ln 10 10sin( x) cos( x) dx. Furthermore, using the chain rule for the second term, d dxqx + x = d.