Calculus, second derivative. Please help! Show work, if possible please. Thanks! :)

y''=sec^3(x)*csc(x)+sec^2(x)*tan(x)-2cot(x)*csc(x)*sec(x)*tan(x)+cot^2(x)*csc(x)*sec(x)+csc^3(x)*sec(x)

Step-by-step explanation:

we are given y= csc x sec x

We use the rule for multiplying derivatives and the knowledge that the derivative of csc is -csc*cot and the derivative of sec is sec*tan to get

csc(x)*sec(x)*tan(x)+sec(x)*-csc(x)*cot(x)

which simplifies to

csc(x)*sec(x)*tan(x)-sec(x)*csc(x)*cot(x)

Using the same rules again (in tedium so they are not listed but can be inferred) we find the second derivative is

sec^3(x)*csc(x)+sec^2(x)*tan(x)-2cot(x)*csc(x)*sec(x)*tan(x)+cot^2(x)*csc(x)*sec(x)+csc^3(x)*sec(x)

10/60

step-by-step explanation:

if you divide 10 by 60 the answer is 0.166666 while if you divide or solve the other problems it equals 0.6

168

step-by-step explanation:

i am pretty sure that the answer is 168