How do you use implicit differentiation to find $\dfrac{{{d^2}y}}{{d{x^2}}}$ of ${x^3} + {y^3} = 1$?

Hint:In implicit differentiation, we have to differentiate each side of the equation without bringing it in a standard form. Here, concepts of differentiation will be used such as chain rule i.e.,

$\dfrac{d}{{dx}}f(g(x)) = f'(g(x)) \times g'(x)$ and quotient rule of differentiation i.e.,
$\dfrac{d}{{dx}}\left[ {\dfrac{{f(x)}}{{g(x)}}} \right] = \dfrac{{f'(x)g(x) - f(x)g'(x)}}{{{{\left[ {g(x)}
\right]}^2}}}$. After using these formulae, we need to simplify the expression such that it becomes in terms of the given variables $x$ and $y$ only.

Complete step by step solution:
(i)As we are asked to differentiate implicitly, we have to differentiate each side of an equation with two variables by treating one of the variables as a function of the other. So, here we have to use the chain rule, i.e.,
$\dfrac{d}{{dx}}f(g(x)) = f'(g(x)) \times g'(x)$

Since we have to calculate $\dfrac{{{d^2}y}}{{d{x^2}}}$, we will first find $\dfrac{{dy}}{{dx}}$
So, we are given:
${x^3} + {y^3} = 1$

Differentiating both the sides with respect to $x$, we will get:
$3{x^2} + 3{y^2}\dfrac{{dy}}{{dx}} = 0$

If we subtract $3{x^2}$ from both the sides, it will become:
$3{y^2}\dfrac{{dy}}{{dx}} = - 3{x^2}$

After this we will divide both the sides by $3{y^2}$, it will become:
$\dfrac{{dy}}{{dx}} = \dfrac{{ - 3{x^2}}}{{3{y^2}}}$

On simplifying it further, we will get:
$\dfrac{{dy}}{{dx}} = \dfrac{{ - {x^2}}}{{{y^2}}}$

(ii)Now let us find $\dfrac{{{d^2}y}}{{d{x^2}}}$.

As we have,
$\dfrac{{dy}}{{dx}} = \dfrac{{ - {x^2}}}{{{y^2}}}$
Again, differentiating both the sides with respect to $x$ we will obtain $\dfrac{{{d^2}y}}{{d{x^2}}}$.

Therefore, $\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {\dfrac{{ - {x^2}}}{{{y^2}}}} \right)$

Here, we have to apply the quotient rule for differentiation, i.e.,
$\dfrac{d}{{dx}}\left[ {\dfrac{{f(x)}}{{g(x)}}} \right] = \dfrac{{f'(x)g(x) - f(x)g'(x)}}{{{{\left[ {g(x)}
\right]}^2}}}$

Therefore, applying the quotient rule here, we get:

$\dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{{\left( {2x} \right)\left( {{y^2}} \right) - \left( {{x^2}} \right)\left(
{2y \times \dfrac{{dy}}{{dx}}} \right)}}{{{{\left( {{y^2}} \right)}^2}}}$

Taking out the common $2x$ from the numerator, we will get:
$\dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{{2x\left( {{y^2} - xy\dfrac{{dy}}{{dx}}} \right)}}{{{y^4}}}$

Since, we know that $\dfrac{{dy}}{{dx}} = \dfrac{{ - {x^2}}}{{{y^2}}}$, we will substitute

$\dfrac{{dy}}{{dx}}$ as $\dfrac{{ - {x^2}}}{{{y^2}}}$ in the above equation:
$\dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{{2x\left[ {{y^2} - xy\left( { - \dfrac{{{x^2}}}{{{y^2}}}} \right)}
\right]}}{{{y^4}}}$
i.e.,
$\dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{{2x\left[ {{y^2} - x\left( { - \dfrac{{{x^2}}}{y}} \right)}
\right]}}{{{y^4}}}$

To simplify it further, we will multiply the numerator and denominator by $y$,
$\dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{{2xy\left[ {{y^2} - x\left( { - \dfrac{{{x^2}}}{y}} \right)}
\right]}}{{{y^5}}}$

Now, multiplying $y$ inside the parentheses in numerator, we will get:
$\dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{{2x\left( {{y^3} + {x^3}} \right)}}{{{y^5}}}$

As we are given in the question, ${x^3} + {y^3} = 1$, we will substitute the values:
$\dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{{2x}}{{{y^5}}}$

Hence $\dfrac{{{d^2}y}}{{d{x^2}}}$ of ${x^3} + {y^3} = 1$ is $ - \dfrac{{2x}}{{{y^5}}}$

Note: When we will calculate the double derivative of the expression by differentiating its first derivative, again a term $\dfrac{{dy}}{{dx}}$ will come. We need to substitute this term with the value we calculated equivalent to it so that the final expression of $\dfrac{{{d^2}y}}{{d{x^2}}}$ does not contain $\dfrac{{dy}}{{dx}}$ in its expression.