If y=cos ^{-1}left(frac{6 x-2 x^2-4}{2 x^2-6 | vedclass

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(D) Given $y=\cos ^{-1}\left(\frac{6 x-2 x^2-4}{2 x^2-6 x+5}\right)$.Let $v = \frac{6 x-2 x^2-4}{2 x^2-6 x+5}$.Note that $v = \frac{-(2x^2-6x+5)+1}{2x

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$\frac{2}{2 x^2-6 x+5}$

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(D) Given $y=\cos ^{-1}\left(\frac{6 x-2 x^2-4}{2 x^2-6 x+5}\right)$.Let $v = \frac{6 x-2 x^2-4}{2 x^2-6 x+5}$.Note that $v = \frac{-(2x^2-6x+5)+1}{2x^2-6x+5} = -1 + \frac{1}{2x^2-6x+5}$.Using the chain rule, $\frac{dy}{dx} = \frac{-1}{\sqrt{1-v^2}} \cdot \frac{dv}{dx}$.Calculating $\frac{dv}{dx}$ using the quotient rule:$\frac{dv}{dx} = \frac{(6-4x)(2x^2-6x+5) - (6x-2x^2-4)(4x-6)}{(2x^2-6x+5)^2} = \frac{2(3-2x)(2x^2-6x+5) + 2(2x^2-6x+4)(2x-3)}{(2x^2-6x+5)^2} = \frac{2(2x-3)[-(2x^2-6x+5) + (2x^2-6x+4)]}{(2x^2-6x+5)^2} = \frac{2(2x-3)(-1)}{(2x^2-6x+5)^2} = \frac{-2(2x-3)}{(2x^2-6x+5)^2}$.Now, $1-v^2 = 1 - \left(\frac{6x-2x^2-4}{2x^2-6x+5}\right)^2 = \frac{(2x^2-6x+5)^2 - (6x-2x^2-4)^2}{(2x^2-6x+5)^2} = \frac{(2x^2-6x+5 - 6x+2x^2+4)(2x^2-6x+5 + 6x-2x^2-4)}{(2x^2-6x+5)^2} = \frac{(4x^2-12x+9)(1)}{(2x^2-6x+5)^2} = \frac{(2x-3)^2}{(2x^2-6x+5)^2}$.Thus, $\sqrt{1-v^2} = \frac{|2x-3|}{2x^2-6x+5}$.Substituting back: $\frac{dy}{dx} = \frac{-1}{\frac{|2x-3|}{2x^2-6x+5}} \cdot \frac{-2(2x-3)}{(2x^2-6x+5)^2} = \frac{2(2x-3)}{|2x-3|(2x^2-6x+5)}$.Assuming $2x-3 > 0$, $\frac{dy}{dx} = \frac{2}{2x^2-6x+5}$.

If $y=\cos ^{-1}\left(\frac{6 x-2 x^2-4}{2 x^2-6 x+5}\right)$, then $\frac{d y}{d x}=$

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