How do you find the second derivative by implicit differentiation?

Hint: In order to determine the second derivative by implicit function, we will consider an implicit function, $ {x^3} + {y^3} = 1 $ . Then, we will find the first order derivative by differentiating the implicit function with respect to $ x $ . And, again differentiate the first order derivative to determine the second order derivative of the implicit function. Thus, by evaluating it we will get the required value.

Complete step-by-step answer:
Now, we need to determine the second derivative by implicit function.
Let us consider an implicit function and determine the second derivative of that function.
 $ {x^3} + {y^3} = 1 $ $ \to \left( 1 \right) $
Now, let us differentiate the equation $ \left( 1 \right) $ with respect to $ x $ , in order to determine the first derivative.
  $ \dfrac{d}{{dx}}\left( {{x^3} + {y^3}} \right) = \dfrac{d}{{dx}}\left( 1 \right) $
We know that $ \dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} $ and $ \dfrac{d}{{dx}}\left( C \right) = 0 $ where C is constant.
Therefore, we have,
 $ 3{x^2} + 3{y^2}\dfrac{{dy}}{{dx}} = 0 $
Let us bring $ \dfrac{{dy}}{{dx}} $ in one side of the equation and the other terms to the other side of the equation.
 $ 3{y^2}\dfrac{{dy}}{{dx}} = - 3{x^2} $
 $ \dfrac{{dy}}{{dx}} = - \dfrac{{3{x^2}}}{{3{y^2}}} $
 $ \dfrac{{dy}}{{dx}} = - \dfrac{{{x^2}}}{{{y^2}}} $ $ \to \left( 2 \right) $
 Now, let us differentiate the equation $ \left( 2 \right) $ with respect to $ x $ , in order to determine the second derivative.
 $ \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( { - \dfrac{{{x^2}}}{{{y^2}}}} \right) $
We know that $ \dfrac{{dy}}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}} $
 $ \dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{{{y^2}\dfrac{d}{{dx}}\left( {{x^2}} \right) - {x^2}\dfrac{d}{{dx}}\left( {{y^2}} \right)}}{{{{\left( {{y^2}} \right)}^2}}} $
 $ \dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{{{y^2}.2x - {x^2}.2y\dfrac{{dy}}{{dx}}}}{{{y^4}}} $
 $ \dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{{2x{y^2} - 2{x^2}y\dfrac{{dy}}{{dx}}}}{{{y^4}}} $
 $ \dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{{2x\left( {{y^2} - xy\dfrac{{dy}}{{dx}}} \right)}}{{{y^4}}} $
From equation $ \left( 2 \right) $ , we know that $ \dfrac{{dy}}{{dx}} = - \dfrac{{{x^2}}}{{{y^2}}} $ .
Therefore, by substituting, we get,
 $ \dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{{2x\left( {{y^2} - xy\left( { - \dfrac{{{x^2}}}{{{y^2}}}} \right)} \right)}}{{{y^4}}} $
 $ \dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{{2x\left( {{y^2} + \dfrac{{{x^3}}}{y}} \right)}}{{{y^4}}} $
 $ \dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{{\dfrac{{2x}}{y}\left( {{y^3} + {x^3}} \right)}}{{{y^4}}} $
 $ \dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{{2x\left( {{y^3} + {x^3}} \right)}}{{{y^5}}} $
Now, from the equation $ \left( 1 \right) $ , we know that $ {x^3} + {y^3} = 1 $ .
Therefore, by substituting, we get,
 $ \dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{{2x}}{{{y^5}}} $
Therefore, the second derivative of the implicit function $ {x^3} + {y^3} = 1 $ is $ - \dfrac{{2x}}{{{y^5}}} $

Note: Functions in which the dependent variable and independent variable are not separated on opposite sides of the equality are known as implicit function. To differentiate an implicit function, differentiate the equation with respect to $ x $ . Collect all the $ \dfrac{{dy}}{{dx}} $ on one side and solve for $ \dfrac{{dy}}{{dx}} $ .
Differentiating helps us to determine the rates of change. There are a number of simple rules which can be used to allow us to differentiate many functions easily.