How do you differentiate # g(x) = (cotx + cscx)/(tanx - sinx) #?

1 Answer
Jan 1, 2018

#(dg)/(dx)=-cotxcot^2(x/2)(csc(x/2)+cscx)#

Explanation:

Quotient rule states if #g(x)=(a(x))/(b(x))#

then #(dg)/(dx)=((da)/(dx)xxb(x)-(db)/(dx)xxa(x))/(b(x))^2#

Here #a(x)=cotx+cscx# and #b(x)=tanx-sinx#

i.e. #(da)/(dx)=-csc^2x-cotxcscx# and #(db)/(dx)=sec^2x-cosx#

Hence #(dg(x))/(dx)=([(tanx-sinx)(-csc^2x-cotxcscx)-(sec^2x-cosx)(cotx+cscx)])/(tanx-sinx)^2#

= #([-tanxcsc^2x-cscx+cscx+cotx -secxcscx-sec^2xcscx+cosxcotx+cotx])/(tanx-sinx)^2#

= #([-tanxcsc^2x+2cotx -secxcscx-sec^2xcscx+cosxcotx])/(tanx-sinx)^2#

We can simplify it further by converting each trigonometric ratio into #sinx# and #cosx# and simplifying it further

but the simpler way could be that as #g(x)=(cotx+cscx)/(tanx-sinx)#

= #(cosx/sinx+1/sinx)/(sinx/cosx-sinx)#

= #((cosx+1)/sinx)/((sinx-sinxcosx)/(sinxcosx)#

= #(cosx(1+cosx))/(sinx(1-cosx))#

=#cotxcot^2(x/2)#

and hence #(dg)/(dx)=2cotxcot(x/2)(-cot(x/2)csc(x/2))xx1/2+cot^2(x/2)(-cotxcscx)#

= #-cotxcot^2(x/2)csc(x/2)-cotxcscxcot^2(x/2)#

= #-cotxcot^2(x/2)(csc(x/2)+cscx)#