Question

If $x^3-2x^2{y}^2+5x+y-5=0$ and $y\left(1\right)=1$ then

• A. $y^{\prime }\left(1\right)=\frac{4}{3}$
• B. $y^{\prime \prime }\left(1\right)=-\frac{4}{3}$
• C. $y^{\prime \prime }\left(1\right)=-8\frac{22}{27}$
• D. $y^{\prime }\left(1\right)=\frac{2}{3}$

Answer 1

Answer: (A), (C)

The correct options are
A $y^{\prime }\left(1\right)=\frac{4}{3}$
C $y^{\prime \prime }\left(1\right)=-8\frac{22}{27}$

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

derivative of equation w.r.t. $x$

$\Rightarrow 3x^2-4x{y}^2-4x^{2}y{y}^{\prime }+5+y^{\prime }-0=0-----\left(1\right)$

Given that $y\left(1\right)=1$

$\Rightarrow 3-4-4y^{\prime }\left(1\right)+5+y^{\prime }\left(1\right)=0\Rightarrow y^{\prime }\left(1\right)=\frac{4}{3}$

Derivative of $\left(1\right)$ w.r.t. $x$

$\Rightarrow 6x-4y^2-8xy{y}^{\prime }-8xy{y}^{\prime }-4x^2{y}^{\prime }{y}^{\prime }-4x^{2}y{y}^{\prime \prime }+y^{\prime \prime }=0$

$\Rightarrow 6-4-8\times \frac{4}{3}-8\times \frac{4}{3}-4\times \frac{16}{9}-4y^{\prime \prime }\left(1\right)+y^{\prime \prime }\left(1\right)=0$

$\Rightarrow y^{\prime \prime }\left(1\right)=\frac{-238}{27}=-8\frac{22}{27}$