How do you differentiate #f(x)=(cotx)/(1+cotx)#?

2 Answers
1s2s2p
Jan 22, 2018

#f'(x)=-(csc^2x)/(1+cotx)^2#

Explanation:

#f(x)=cotx/(1+cotx)#

#f'(x)=d/dx[cotx/(1+cotx)]#

#color(white)(f'(x))=((1+cotx)d/dx[cotx]-cotxd/dx[1+cotx])/(1+cotx)^2#

#color(white)(f'(x))=((1+cotx)(-csc^2x)-cotx(-csc^2x))/(1+cotx)^2#

#color(white)(f'(x))=(-csc^2x(1+cotx)+cotxcsc^2x)/(1+cotx)^2#

#color(white)(f'(x))=(-csc^2x(1+cotx-cotx))/(1+cotx)^2#

#color(white)(f'(x))=(-csc^2x(1))/(1+cotx)^2#

#color(white)(f'(x))=(-csc^2x)/(1+cotx)^2#

#color(white)(f'(x))=-(csc^2x)/(1+cotx)^2#

Jim G.
Jan 22, 2018

#f'(x)=-(csc^2x)/(1+cotx)^2#

Explanation:

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

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

#f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larrcolor(blue)"quotient rule"#

#g(x)=cotxrArrg'(x)=-csc^2x#

#h(x)=1+cotxrArrh'(x)=-csc^2x#

#rArrf'(x)=((1+cotx)(-csc^2x)+cotxcsc^2x)/(1+cotx)^2#

#color(white)(rArrf'(x))=(-csc^2xcancel(-cotxcsc^2c)cancel(+cotxcsc^2x))/(1+cotx)^2#

#=-(csc^2x)/(1+cotx)^2#

Related questions
See all questions in Derivatives of y=sec(x), y=cot(x), y= csc(x)
Impact of this question
15261 views around the world
You can reuse this answer
Creative Commons License