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

1 Answer
Jan 23, 2018

Answer:

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

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

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

Explanation:

We are given #f(x)=(e^tan(1/x))^(1/2)#

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

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

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

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

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

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

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

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