How do you differentiate #tan(2pix+pi/2)#?
1 Answer
May 6, 2018
Explanation:
#"differentiate using the "color(blue)"chain rule"#
#"given "y=f(g(x))" then"#
#dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"#
#"here "y=tan(2pix+pi/2)#
#rArrdy/dx=sec^2(2pix+pi/2)xxd/dx(2pix+pi/2)#
#color(white)(rArrdy/dx)=2pisec^2(2pix+pi/2)#
