How do you differentiate #f(x)=1/sqrtsec(e^(x) ) # using the chain rule?

1 Answer
Jan 30, 2018

Answer:

#f'(x)=-(e^xtan(e^x))/(2(sec(e^x))^(1/2))#

#color(white)(f'(x))=-(e^xtan(e^x)sec(e^x)^(-1/2))/2#

Explanation:

Here we have #f(x)=(sec(e^x))^(-1/2)#

#f'(x)=d/dx[(sec(e^x))^(-1/2)]#

#color(white)(f'(x))=(sec(e^x))^(-1/2-1)*-1/2*d/dx[sec(e^x)]#

#color(white)(f'(x))=(sec(e^x))^(-3/2)*-1/2*d/dx[sec(e^x)]#

#color(white)(f'(x))=(sec(e^x))^(-3/2)*-1/2*sec(e^x)tan(e^x)*d/dc[e^x]#

#color(white)(f'(x))=-(e^xsec(e^x)tan(e^x))/(2(sec(e^x))^(3/2))#

#color(white)(f'(x))=-(e^xtan(e^x))/(2(sec(e^x))^(1/2))#

#color(white)(f'(x))=-(e^xtan(e^x)sec(e^x)^(-1/2))/2#