How do you use the chain rule to differentiate #y=sec2x^4#?
1 Answer
Jan 19, 2018
Explanation:
#"given "y=g(h(x))" then "#
#dy/dx=g'(h(x))xxh'(x)#
#y=sec(2x^4)#
#rArrdy/dx=sec(2x^4)tan(2x^4)xxd/dx(2x^4)#
#color(white)(rArrdy/dx)=8x^3sec(2x^4)tan(2x^4)#
