Given 3xy2+y4x−(2x2−1)(y−5)=11...(1)To find dxdy,when x=1.We use implicit differentiationSubstituting x=1 in (1),we get3(1)y2+y4(1)−(2(1)2−1)(y−5)=11⇒3y2+y4−(y−5)=11⇒y3y3+4−y(y−5)=11⇒3y3+4−y(y−5)=11y (by cross multiplication)⇒3y3+4−y2+5y−11y=0⇒3y3−y2−6y+4=0 Since the sum of the coefficients is equal to zeroy=1 is a root of 3y3−y2−6y+4=0By synthetic division we get 3y2+2y−4 as a factor⇒y=3−1+(13) and y=3−1−(13)∴y=1, 3−1+(13) and 3−1−(13)Differentiating (1), with \ respect \ to \ 'x',\ we \ get \\
\frac{d}{dx}(3x^2y +\frac{4x}{y}-\frac{(y-5)})=\frac{d}{dx}(11) \\
\Rightarrow
\frac{d}{dx}(3x^2y)+\frac{d}{dx}(\frac{4x}{y})-\frac{d}{dx}(\frac{(y-5)}{(2x^2-1)})=\frac{d}{dx}(11)\\
\Rightarrow [(3x^2)\frac{d}{dx}(y) +3y\frac{d}{dx}(x^2)]+[\frac{y\frac{d}{dx}(4x)-4x\frac{d}{dx}(y)}{y^2}]\\
-[\frac{(2x^2-1)\frac{d}{dx}(y-5)-(y-5)\frac{d}{dx}(2x^2-1)}{(2x^2-1)^2}]=0\\
\Rightarrow [(3x^2)\frac{dy}{dx} +6xy]+[\frac{4y-4x\frac{dy}{dx}}{y^2}]\\
-[\frac{(2x^2-1)\frac{dy}{dx}-(y-5)(4x)}{(2x^2-1)^2}]=0\\
Regrouping \ the \ terms \ containing \ \frac{dy}{dx}, we get \\
[3x^2-\frac{4x}{y^2}-\frac{1}{(2x^2-1)}]\frac{dy}{dx}=[-6xy-\frac{4y}{y^2}-\frac{(y-5)(4x)}{(2x^2-1)^2}]\\
\Rightarrow \frac{dy}{dx}=\frac{[-6xy-\frac{4y}{y^2}-\frac{(y-5)(4x)}{(2x^2-1)^2}]}{[3x^2-\frac{4x}{y^2}-\frac{1}{(2x^2-1)}]}\\
Substituting \ x=1 \ and \ y=1 \ in \frac{dy}{dx}, we \ get \\
[\frac{dy}{dx}]_{x=1,y=1}=\frac{[-6(1)(1)-\frac{4(1)}{(1)^2}-\frac{(1-5)(4(1))}{(2(1)^2-1)^2}]}{[3(1)^2-\frac{4(1)}{(1)^2}-\frac{1}{(2(1)^2-1)}]}\\
=\frac{[-6-4+16]}{[3-4-1]}\\
=\frac{(6)}{(-2)}=-3\\
\therefore [\frac{dy}{dx}]_{x=1,y=1}=-3
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