Question

The value of $y^{\prime \prime }\left(1\right)$ if $x^3-2x^2{y}^2+5x+y-5=0$ when $y\left(1\right)=1$, is equal to

• A. $\frac{22}{7}$
• B. $-7\frac{21}{28}$
• C. $8$
• D. $-8\frac{22}{27}$

Answer 1

The correct option is D $-8\frac{22}{27}$

Differentiating the given expression, we get
$3x^2-4x{y}^2-4x^{2}y{y}^{\prime }+5+y^{\prime }=0$
Putting x = 1, we have
$3.1-\mathrm{4.1.1}-\mathrm{4.1.1}{y}^{\prime }\left(1\right)+5+y^{\prime }\left(1\right)=0$
$\Rightarrow 4-3y^{\prime }\left(1\right)=0\Rightarrow y^{\prime }\left(1\right)=\frac{4}{3}$
Differentiating again, we have
$6x-4y^2-8xy{y}^{\prime }-8xy{y}^{\prime }-4x^2(y^{\prime }{)}^2-4x^{2}y{y}^{\prime \prime }+y^{\prime \prime }=0$
Putting x = 1, y = 1 and $y^{\prime }\left(1\right)=\frac{4}{3}$, we get
$6-4-8\left(\frac{4}{3}\right)-8\left(\frac{4}{3}\right)-4\left(\frac{16}{9}\right)-3y^{\prime \prime }\left(1\right)=0$
$\Rightarrow y^{\prime \prime }\left(1\right)=-8\frac{22}{27}$