Question

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

• A. $y^{\text{'}}\left(1\right)=\frac{2}{3}$
• B. $y^{\text{'}}\left(1\right)=\frac{4}{3}$
• C. $y^{\text{'}\text{'}}\left(1\right)=-8\left(\frac{22}{7}\right)$
• D. $y^{\text{'}}\left(1\right)=-\frac{4}{3}$

Answer 1

Answer: (B), (C)

$y^{\text{'}\text{'}}\left(1\right)=-8\left(\frac{22}{7}\right)$

Explanation for the correct options:

Step-1: Derivative of first order:

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

Derivative of the given equation w.r.t $x$ will be

$\begin{array}{rcl}3x^2-2\left[x^2·2yy\text{'}+y^2·2x\right]+5+y^{\text{'}}& =& 0\\ \Rightarrow 3x^2-4x{y}^2-4x^{2}yy\text{'}+5+y^{\text{'}}& =& 0.......\left(1\right)\end{array}$

By substituting $y\left(1\right)=1$, we get

$\begin{array}{rcl}3-4-4y^{\text{'}}+5+y^{\text{'}}& =& 0\\ \Rightarrow y^{\text{'}}\left(1\right)& =& \frac{4}{3}\end{array}$

Hence, option B is correct.

Step-2: Derivative of second order:

$\begin{array}{rcl}3x^2-4x{y}^2-4x^{2}yy\text{'}+5+y^{\text{'}}& =& 0\end{array}$

Now let us find $y^{\text{'}\text{'}}$. For this we will find the derivative of $\left(1\right)$ w.r.t $x$, and it will be

$6x-4y^2-8xy{y}^{\text{'}}-8xyy\text{'}-4x^2{y}^{\text{'}}{y}^{\text{'}}-4x^{2}y{y}^{\text{'}\text{'}}+y^{\text{'}\text{'}}=0$

By substituting $y\left(1\right)=1\mathrm{and}y\text{'}\left(1\right)=\frac{4}{3}$, we get

$\begin{array}{rcl}6-4-8\left(\frac{4}{3}\right)-8\left(\frac{4}{3}\right)-4{\left(\frac{4}{3}\right)}^2-4y^{\text{'}\text{'}}+y^{\text{'}\text{'}}& =& 0\\ \Rightarrow -\frac{238}{9}& =& 3y\text{'}\text{'}\\ \Rightarrow y\text{'}\text{'}\left(1\right)& =& -8\frac{22}{27}\end{array}$

Hence, option B and option C are correct.