If x³-2 x² y²+5 x+y-5=0, then at (1,1) y''(1)=

Correct Answer is : −23827-238/2727−238

Correct Answer is : −23827-238/2727−238

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AP EAPCET 04-Jul-2022 Shift 1 Paper

Section: Mathematics

Question : 64 of 160

Marks: +1, -0

x^{3}-2 x^{2} y^{2}+5 x+y-5=0

y''(1)=

Solution: 👈: Video Solution

x^3-2 x^2 y^2+5 x+y-5=0

On differentiating w.r.t.

x

, we get

3 x^2-4 x y^2-4 x^2 y \cdot y'+5+y'-0=0

........(i)

On putting

x=1, y=1

\;3-4-4 y'+5+y'=0

\;\Rightarrow\;\; y'=\;\frac{4}{3}

\;y'(1)=\;\frac{4}{3}

On differentiating Eq. (i) w.r.t.

x

6 x-4 y^2-8 x y y'-8 x y y'-4 x^2 y' \cdot y'-4 x^2 y y''+y''=0

On putting

x=1, y=1

and

y'=\;\frac{4}{3}

to obtain

y''(1)

\; \Rightarrow 6-4-8 \times \; \frac{4}{3}-8 \times \; \frac{4}{3}-4 \times \; \frac{4}{3} \times \; \frac{4}{3}-4 \times y''+y''=0

\; \Rightarrow 2-\;\frac{64}{3}-\;\frac{64}{9}-3 y''=0

\Rightarrow \; \frac{-238}{9}=3 y''

\Rightarrow y''=-\; \frac{238}{27}

\; \text{ or } \;\; y''(1)=-\; \frac{238}{27}

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