How do you differentiate #f(x)=(1+e^(2x))/(2-e^(2x))#?

1 Answer
Jim G.
Feb 25, 2017

#f'(x)=(6e^(2x))/(2-e^(2x))^2#

Explanation:

differentiate using the #color(blue)"quotient rule"#

#"Given "f(x)=(g(x))/(h(x))" then"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2)color(white)(2/2)|)))#

#"here "g(x)=1+e^(2x)rArrg'(x)=e^(2x).d/dx(2x)=2e^(2x)#

#"and "h(x)=2-e^(2x)rArrh'(x)=-e^(2x).d/dx(2x)=-2e^(2x)#

#rArrf'(x)=((2-e^(2x))2e^(2x)-(1+e^(2x))(-2e^(2x)))/(2-e^(2x))^2#

#color(white)(xxxxxxx)=(4e^(2x)cancel(-2e^(4x))+2e^(2x)cancel(+2e^(4x)))/(2-e^(2x))^2#

#color(white)(xxxxxxx)=(6e^(2x))/(2-e^(2x))^2#

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