Question

help :) with derivatives

Answer 1

u can use direct approach like
\[\frac{ d }{ dx }\cot (ax)=-a*cosec^2(ax)\]
and
\[\frac{ d }{ dx }cosec(ax)=-a*cosec(ax).\cot(ax)\]

Answer 2

or u can break them in sin and cos then use product or division rule

Answer 3

ok?

Answer 4

so consider the first term cot(2x)
so
\[\frac{ d }{ dx }\cot(2x)=-2cosec^2(2x)\]

Answer 5

correct?

Answer 6

and for the second term\[\frac{ d }{ dx }cosec(2x)=-2cosec(2x).\cot(2x)\]
now add them up-\[so \\K'(x)=-2cosec^2(2x)-2cosec(2x).\cot(2x)\\=-2cosec(2x)[cosec(2x)+\cot(2x)]\]
yup u're correct :)

Answer 7

got it then ?