Question

find the derivative of
y=2cotx+sec3x+cscX

find d^2/dx^2 for 1+y=x+xy

Answer 1

The derivatives of these trig functions should be right there in your Calculus text.

for the second of your questions, I will use y' for the first derivative and y" for the second derivative i.e. y'= dy/dx

1+y=x+xy so
y' = 1 + y + xy'
y'(1-x) = 1+y
y' = (1+y)/(1-x)

y" = [(1-x)y' - (1+y)(-1)]/((1-x)^2)
=[(1+y) + (1+y)]/((1-x)^2) after I replaced y'
=(2+2y)/((1-x)^2)