#(Sinx-cosx)/(sinx+cosx)# Show that #(dy)/(dx)=1+y^2?#

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Jim G. Share
Jan 11, 2018

Answer:

#"see explanation"#

Explanation:

#"let "y=(sinx-cosx)/(sinx+cosx)#

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

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

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

#g(x)=sinx-cosxrArrg'(x)=cosx+sinx#

#h(x)=sinx+cosxrArrh'(x)=cosx-sinx#

#dy/dx=((sinx+cosx)(sinx+cosx)-(sinx-cosx)-1(sinx-cosx))/(sinx+cosx)^2#

#=((sinx+cosx)^2+(sinx-cosx)^2)/(sinx+cosx)^2#

#=(sinx+cosx)^2/(sinx+cosx)^2+(sinx-cosx)^2/(sinx+cosx)^2#

#=1+y^2#

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Write your answer here...
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Answer

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Answer:

Explanation

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Jan 11, 2018

Answer:

My explanation is below.

Explanation:

#y=(sinx-cosx)/(sinx+cosx)#

#dy/dx=[(cosx+sinx)(sinx+cosx)-(cosx-sinx)*(sinx-cosx)]/(sinx+cosx)^2#

=#[(sinx+cosx)^2+(sinx-cosx)^2]/(sinx+cosx)^2#

=#1+[(sinx-cosx)/(sinx+cosx)]^2#

=#1+y^2#

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