How do you differentiate #(-2x^2 -5x-4 )/ (cos2x)# using the quotient rule?

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Answer 1 · 2017-06-08T07:20:00

#(df)/(dx)=-(cos2x(4x+5)-2sin2x(2x^2+5x+4))/(cos^2 2x)#

Quotient rule states if #f(x)=(g(x))/(h(x))#

then #(df)/(dx)=((dg)/(dx)xxh(x)-(dh)/(dx)xxg(x))/(h(x))^2#

Hence, as #f(x)=(-2x^2-5x-4)/(cos2x)#

#(df)/(dx)=(cos2x(-4x-5)-(-2sin2x)(-2x^2-5x-4))/(cos^2 2x)#

= #(-4xcos2x-5cos2x-2sin2x(2x^2+5x+4))/(cos^2 2x)#

= #(-4xcos2x-5cos2x-4x^2sin2x-10xsin2x-8sin2x)/(cos^2 2x)#

= #-(cos2x(4x+5)-2sin2x(2x^2+5x+4))/(cos^2 2x)#