How do you differentiate # g(x) = sin x + 1/2 cot x #?
1 Answer
Jan 5, 2016
Explanation:
The trick here is knowing your trigonometric derivatives.
#d/dx(sinx)=cosx#
#d/dx(cotx)=-csc^2x#
Hence
#g'(x)=cosx-1/2csc^2x#
