Q6) ii

Q6) ii 11:53 o. 00K/s t 4G VoLTE Sayalee Ambulkar 1 day Iii,i oaaa

Dear Student,


$$\begin{gathered} Given : y={\tan }^{-1}\left(\frac{5x+1}{3-x-6x^2}\right) \\ To Find :\frac{dy}{dx} \\ solution : y={\tan }^{-1}\left(\frac{5x-1}{3-x-6x^2}\right) \\ applying \frac{d}{dx} both the sides \\ \Rightarrow \frac{d}{dx}\left(y\right)=\frac{d}{dx}\left\{{\tan }^{-1}\right(\frac{5x-1}{3-x-6x^2}\left)\right\} \\ \Rightarrow \frac{dy}{dx}=\frac{d\left\{{\tan }^{-1}\right({\displaystyle \frac{5x-1}{3-x-6x^2}}\left)\right\}}{d\left({\displaystyle \frac{5x-1}{3-x-6x^2}}\right)}\times \frac{d\left({\displaystyle \frac{5x-1}{3-x-6x^2}}\right)}{dx} [using Chain Rule] \\ \Rightarrow \frac{dy}{dx}=\frac{1}{1+({\displaystyle \frac{5x-1}{3-x-6x^2}}{)}^2}\times \frac{(3-x-6x^2)(5-0)-(5x-1)(-1-12x)}{(3-x-6x^2{)}^2} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{{\displaystyle \frac{(3-x-6x^2{)}^2+(5x-1)}{(3-x-6x^2{)}^2}}}\times \frac{5(3-x-6x^2{)}^2+5x+60x^2-1-12x}{(3-x-6x^2{)}^2} \\ \Rightarrow \frac{dy}{dx}=\frac{\cancel{(3-x-6x^2{)}^2}}{(3-x-6x^2{)}^2+(5x-1)}\times \frac{5(9+x^2+36x^4-6x+12x^3-36x^2)+60x^2-7x-1}{\cancel{(3-x-6x^2{)}^2}} \\ \Rightarrow \frac{dy}{dx}=\frac{45-175x^2+180x^4-30x+60x^3+60x^2-7x-1}{(9+x^2+36x^4-6x+12x^3-36x^2)+(5x-1)} \\ \Rightarrow \frac{dy}{dx}=\frac{180x^4+60x^3-115x^2-37x+44}{36x^4+12x^3-35x^2-x+8} \end{gathered}$$
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