Question

Derive the equation:${x^3} - 2{x^2}{y^2} + 5x + y - 5 = 0$ at${y_{x = 1}} = 1$
(a) $\dfrac{{{d^2}y}}{{d{x^2}}} = - 8\left( {\dfrac{{22}}{7}} \right)$
(b) $\dfrac{{{d^2}y}}{{d{x^2}}} = - 8$
(c) $\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{22}}{7}$
(d) None of the above

Answer 1

Hint : In calculus, the second derivative, or the second order derivative, of a function f is the derivative of the derivative of f. The second derivative is the rate of change of the rate of change of a point at a graph (the "slope of the slope" if you will). This can be used to find the acceleration of an object (velocity is given by the first derivative).

Complete step-by-step answer :
Since, we have the given equation
${x^3} - 2{x^2}{y^2} + 5x + y - 5 = 0$
As a result, solving the equation, first of all derivating the above given equation with respect to the ‘x’ variable, can reach up to a desire output,
$\Rightarrow 3{x^2} - 2\left[ {{x^2}2y\dfrac{{dy}}{{dx}} + {y^2}2x} \right] + 5 + \dfrac{{dy}}{{dx}} - 0 = 0$
Here, we have use the rules of multiplication in derivation $\Rightarrow \dfrac{d}{{dx}}(xy) = x\dfrac{{dy}}{{dx}} + y(1)$ that is first term as it is into derivation of second term plus second term as it is into derivation of first term. Also, the derivation of any constant is always ‘zero’ that is $\Rightarrow \dfrac{d}{{dx}}(5) = 0$ respectively!
Since, $\dfrac{{dy}}{{dx}}$ also be denoted as $y'$,
$\Rightarrow 3{x^2} - 4{x^2}yy' - 4{y^2}x + 5 + y' = 0$ … (i)
Taking $y'$ common in one bracket and remaining terms in another bracket, we get
$\Rightarrow 3{x^2} - 4{y^2}x + 5 - 4{x^2}yy' + y' = 0$
$\Rightarrow \left( {3{x^2} - 4{y^2}x + 5} \right) = y'\left( {4{x^2}y - 1} \right)$
Solving the above equation mathematically, we get
$\Rightarrow y' = \dfrac{{3{x^2} - 4{y^2}x + 5}}{{4{x^2}y - 1}}$
But, we have given that ${y_{x = 1}} = 1$, we get
$\Rightarrow y' = \dfrac{{3 - 4 + 5}}{{4 - 1}}$
$\Rightarrow y' = \dfrac{4}{3}$
Considering equation (i),
$\Rightarrow 3{x^2} - 4{x^2}yy' - 4{y^2}x + 5 + y' = 0$
Now, since for the absolute answer, again derivating (double derivative) of the above equation, we get
$\Rightarrow 6x - 4\left[ {{x^2}yy'' + y'{x^2}y' + 2xyy'} \right] - 4\left[ {{y^2} + 2xyy'} \right] + 0 + y'' = 0$
$\Rightarrow 6x - 4{x^2}yy'' - 4y'{x^2}y' - 8xyy' - 4{y^2} - 8xyy' + y'' = 0$
(Here, in three terms we have divided into two parts i.e. ${x^2}y$ and $y'$ respectively and then substituting in the formula $\dfrac{d}{{dx}}(xy) = x\dfrac{{dy}}{{dx}} + y(1)$ we get the solution)
Again, substituting $y' = \dfrac{4}{3}$ and ${y_{x = 1}} = 1$ in the above equation to achieve a desired answer, we get
$\Rightarrow 6 - 4y'' - 4 \times \dfrac{4}{3} \times \dfrac{4}{3} - 8 \times \dfrac{4}{3} - 4 - 8 \times \dfrac{4}{3} + y'' = 0$
$\Rightarrow 6 - 4y'' - \dfrac{{64}}{9} - \dfrac{{32}}{3} - 4 - \dfrac{{32}}{3} + y'' = 0$
Solve the equation mathematically, we get
$\Rightarrow - 4y'' - \dfrac{{64}}{9} - \dfrac{{64}}{3} + 2 + y'' = 0$
$\Rightarrow - 4y'' - \dfrac{{256}}{9} + 2 + y'' = 0$
Adding and subtracting the equations, we get
$\Rightarrow - 4y'' - \dfrac{{256}}{9} + y'' = 0$
$\Rightarrow - 4y'' + y'' = \dfrac{{238}}{9}$
Hence, the desired answer is
$\Rightarrow - 3y'' = \dfrac{{238}}{9}$
$\Rightarrow y'' = - \dfrac{{238}}{{27}}$
Converting the obtained value in simplest form, we get
$\therefore \Rightarrow y'' = - 8\dfrac{{22}}{7}$
So, the correct answer is “Option B”.

Note : One must remember the concept of derivation, how to differentiate the equation with respect to which variable, etc.? Also, laws of derivation such as $\dfrac{d}{{dx}}(xy) = x\dfrac{{dy}}{{dx}} + y(1)$is the multiplication rule of derivation used here and $\dfrac{{dy}}{{dx}},\dfrac{{{d^2}y}}{{d{x^2}}},...$can be noted as and so on! Derivation of any constant number is always zero. Deriving the equation with the same term is always one$\dfrac{d}{{dx}}(x) = 1$. Algebraic identities play a significant role in solving this problem.