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

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Answer 1 · 2018-01-01T05:51:38

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

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

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

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

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