How do you differentiate # y=sec (x^2/pi - xpi)# using the chain rule?
1 Answer
Nov 17, 2017
Explanation:
#"given "y=f(g(x))" then "#
#dy/dx=f'(g(x))xxg'(x)larr"chain rule"#
#y=sec(x^2/pi-xpi)#
#rArrdy/dx=sec(x^2/pi-xpi)tan(x^2/pi-xpi)xxd/dx(x^2/pi-xpi)#
#=(2/pix-pi)sec(x^2/pi-pix)tan(x^2/pi-pix)#
