How do you differentiate #f(x)= (1 - sin^2x)/cosx # using the quotient rule?

1 Answer

the answer #f'(x)=-sinx#

Explanation:

The common way is

#sin^2(x)+cos^2(x)=1#

#cos^2(x)=1-sin^2(x)#

#f(x)= (1 - sin^2x)/cosx=cos^2(x)/cos(x)=cos(x)#

#f'(x)=-sin(x)#

By using the quotient rule

#y=f(x)/g(x)#

#y'=[g(x)*f'(x)-f(x)*g'(x)]/(g(x))^2#

#f(x)= (1 - sin^2x)/cosx#

#f'(x)=(sin(x)*(1-sin(x)^2)-2*cos(x)^2*sin(x))/cos(x)^2#

#=((sin(x)*(1-sin(x)^2))/cos(x)^2)-2*sin(x)#

= #(sin(x)*cos^2(x))/cos^2(x)-2sinx#

= #sinx-2sinx#

= #-sinx#