1st and 2nd Derivative of y= xtanx?

Answer 1

y = xtan(x)

Use product rule:
dy/dx = x/cos²(x) + tan(x)
= x/cos²(x) + sin(x)/cos(x)
= [x + sin(x)·cos(x)] / cos²(x)

Use product and chain rules:
d²y/dx² = 1/cos²(x) + (-2)·(x/[cos(x)]^3)·(-sin(x)) + 1/cos²(x)
= 2/cos²(x) + [2·x·sin(x)] / [cos(x)]^3
= [2·cos(x) + 2·x·sin(x)] / [cos(x)]^3

Answer 2

y = x tan(x)

f(x) = x, g(x) = tan(x)

[ 1*tan(x) - x*sec²(x) ] / tan²(x)
[ tan(x) - x sec²(x) ] / tan²(x)
1/tan(x) - [ x sec²(x)/ tan²(x) ]
cot(x) - [ x / cos²(x) tan²(x) ]
cot(x) - [ x / sin²(x) ]
cot(x) - x csc²(x) <----- dy/dx

y' = cot(x) - x csc²(x)

"G":cot(x): f(x)= cos(x), g(x)=sin(x)
[ -sin(x)*sin(x) - cos(x)*cos(x) ] / sin²(x)
[ -sin²(x) - cos²(x) ] / sin²(x)
[ -1 (sin²(x) + cos²(x)) ] / sin²(x)
-1 * 1 / sin²(x)
- csc²(x)

"H": -x csc²(x): f(x) = -x, g(x)= csc²(x)

[ -1*csc²(x) - -x*csc²(x)' ] / csc^4(x)
[ -csc²(x) + x*csc²(x)' ] / csc^4(x)

You can finish it from here. Just take y' of csc²(x) and plug it in above.

then it'll be G + H and we know G = - csc²(x) so just plug in the rest. You gotta learn something so just finsh it up. Its 80% complete.