Solutions to Diff Cal Test #2 Practice Problems: 1. Use the product rule to differentiate. ππ(π₯π₯) = cot π₯π₯ (π₯π₯ + 4)2β3 2 ππ β² (π₯π₯) = (β csc 2 π₯π₯)(π₯π₯ + 4)2β3 + (cot π₯π₯) οΏ½ (π₯π₯ + 4)β1β3 οΏ½ 3 2. Use the quotient rule to differentiate. ππ(π₯π₯) = 2π₯π₯ 3 β 4π₯π₯ 2 + 5π₯π₯ β 9 5 sec π₯π₯ ππ β² (π₯π₯) = (5 sec π₯π₯)(6π₯π₯ 2 β 8π₯π₯ + 5) β (2π₯π₯ 3 β 4π₯π₯ 2 + 5π₯π₯ β 9)(5 sec π₯π₯ tan π₯π₯) (5 sec π₯π₯)2 3. Find the derivative of ππ with respect to π₯π₯. ππ(π₯π₯) = 5 sin2 οΏ½οΏ½3 csc(7π₯π₯ 2 β 2π₯π₯)οΏ½ = 5οΏ½sinοΏ½(3 csc(7π₯π₯ 2 β 2π₯π₯))1β2 οΏ½οΏ½ 2 1 ππ β² (π₯π₯) = 10 sin(3 csc(7π₯π₯ 2 β 2π₯π₯))1β2 β cos(3 csc(7π₯π₯ 2 β 2π₯π₯))1β2 β (3 csc(7π₯π₯ 2 β 2π₯π₯))β1β2 2 β (β3 csc(7π₯π₯ 2 β 2π₯π₯) cot(7π₯π₯ 2 β 2π₯π₯)) β (14π₯π₯ β 2) 4. Find the derivative of ππ with respect to π₯π₯. ππ(π₯π₯) = β3π₯π₯ cot(5π₯π₯ 2 + 4π₯π₯) = β(3π₯π₯ ) cot(5π₯π₯ 2 + 4π₯π₯) ππ β² (π₯π₯) = (β3π₯π₯ ln 3)(cot(5π₯π₯ 2 + 4π₯π₯)) + (β3π₯π₯ )(β csc 2 (5π₯π₯ 2 + 4π₯π₯))(10π₯π₯ + 4) 5. Find the derivative of ππ with respect to π₯π₯. ππ(π₯π₯) = log 2 cos(5π₯π₯) ππ β² (π₯π₯) = (β sin 5π₯π₯)(5) (ln 2)(cos 5π₯π₯) 6. Find the derivative of ππ with respect to π₯π₯. ππ(π₯π₯) = arcsin(π₯π₯ 2 β 7π₯π₯) ππ β² (π₯π₯) = 2π₯π₯ β 7 οΏ½1 β (π₯π₯ 2 β 7π₯π₯)2 7. Find the derivative of ππ with respect to π₯π₯. ππ(π₯π₯) = ππ arctan(2π₯π₯+5) ππ β² (π₯π₯) = ππ arctan(2π₯π₯+5) β 2 1 + (2π₯π₯ + 5)2 8. Find the derivative of ππ with respect to π₯π₯. ππ(π₯π₯) = arccsc π₯π₯ ln(tan(2π₯π₯)) ππ β² (π₯π₯) =οΏ½ β1 |π₯π₯|βπ₯π₯ 2 β 1 οΏ½ (ln(tan 2π₯π₯)) + (arccsc π₯π₯) οΏ½ (sec 2 2π₯π₯)(2) οΏ½ tan 2π₯π₯
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