How do you use the chain rule to differentiate #y=tan(-2x)#?
1 Answer
May 8, 2017
Explanation:
#"Given " y=f(g(x))" then"#
#dy/dx=f'(g(x))xxg'(x)larr" chain rule"#
#color(orange)"Reminder " d/dx(tanx)=sec^2x#
#f(g(x))=tan(-2x)rArrf'(g(x))=sec^2(-2x)#
#g(x)=-2xrArrg'(x)=-2#
#rArrdy/dx=-2sec^2(-2x)#
