How do you differentiate #f(x)=sec^2(e^(x^4) ) # using the chain rule?

1 Answer

Answer:

#f^'(x)=8x^3(e^(x^(4)))sec^2(e^(x^(4)))tan(e^(x^(4)))#

Explanation:

Chain Rule:
#color(red)(d/(dx)(g@f(x))=g^'(f(x))*f^'(x))#
#f(x)=sec^2(e^(x^(4)))=[color(red)(sec(e^(x^(4))))]^2#
#f^'(x)=2sec(e^(x^(4)))*d/(dx)color(red)((seccolor(blue)((e^(x^(4)))))#
#f^'(x)=2sec(e^(x^(4)))sec(e^(x^(4)))tan(e^(x^(4)))d/(dx)color(blue)((e^(x^4))#
#f^'(x)=2sec(e^(x^(4)))sec(e^(x^(4)))tan(e^(x^(4)))(e^(x^(4)))d/(dx)(color(red)(x^4))#
#f^'(x)=2sec(e^(x^(4)))sec(e^(x^(4)))tan(e^(x^(4)))(e^(x^(4)))(4x^3)#
#f^'(x)=8x^3(e^(x^(4)))sec(e^(x^(4)))sec(e^(x^(4)))tan(e^(x^(4)))#
#f^'(x)=8x^3(e^(x^(4)))sec^2(e^(x^(4)))tan(e^(x^(4)))#