Question

How do you find \[{y}''\] by implicit differentiation of \[{{x}^{3}}+{{y}^{3}}=1\]?

Answer 1

Hint: In an implicit differentiation, we differentiate the given equation which has two variables such that one of the variables is treated as the function of the other and then differentiated. The second derivative of a function can also be written as \[{y}''\].

Complete step by step solution:
As per the given question, we need to find \[{y}''\] by implicitly differentiating \[{{x}^{3}}+{{y}^{3}}=1\].
To find \[{y}''\] first we need to \[{y}'\].
We know that according to power rule, \[\dfrac{d}{dx}\left( a{{x}^{n}} \right)=\left( na \right){{x}^{n-1}}\].
We know that the derivative of a constant is 0.
Let us assume y is a function of x then we differentiate the equation.
Now we differentiate the equation implicitly
\[\Rightarrow {{x}^{3}}+{{y}^{3}}=1\]
Since both x and y have powers, we use power rule to differentiate the equation.
\[\Rightarrow 3{{x}^{2}}+3{{y}^{2}}\dfrac{dy}{dx}=0\]
We know that \[\dfrac{dy}{dx}\] can also be written as \[{y}'\].
Then the equation becomes
\[\Rightarrow 3{{x}^{2}}+3{{y}^{2}}{y}'=0\]
\[\Rightarrow {{x}^{2}}+{{y}^{2}}{y}'=0\]
Here to differentiate \[{{y}^{2}}{y}'\] we need to use the product rule. The rule follows from the limit definition of derivative and is given by \[\dfrac{dy}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\] if \[y=uv\].
According to power rule, the derivative of \[a{{x}^{n}}\] is given by \[\dfrac{d}{dx}\left( a{{x}^{n}} \right)=\left( na \right){{x}^{n-1}}\].
Using this product rule, the derivative of \[{{y}^{2}}{y}'\] will be \[{{y}^{2}}{y}''+{y}'\left( 2y{y}' \right)={{y}^{2}}{y}''+2y{{\left( {{y}'} \right)}^{2}}\].
And the derivative of \[{{x}^{2}}\] is \[2x\]. Now the equation will be
\[\Rightarrow 2x+{{y}^{2}}{y}''+2y{{\left( {{y}'} \right)}^{2}}=0\]
The value of \[{y}'\] from the equation \[{{x}^{2}}+{{y}^{2}}{y}'=0\] is \[{y}'=\dfrac{-{{x}^{2}}}{{{y}^{2}}}\] on substituting this value in above equation. We get
\[\begin{align}
  & \Rightarrow 2x+{{y}^{2}}{y}''+{{\left( \dfrac{-{{x}^{2}}}{{{y}^{2}}} \right)}^{2}}\left( 2y \right)=0 \\
 & \Rightarrow 2x+{{y}^{2}}{y}''+\left( \dfrac{2{{x}^{4}}}{{{y}^{3}}} \right)=0 \\
\end{align}\]
\[\Rightarrow 2x{{y}^{3}}+{{y}^{5}}{y}''+2{{x}^{4}}=0\]
\[\Rightarrow {{y}^{5}}{y}''=-2{{x}^{4}}-2x{{y}^{3}}\]
\[\Rightarrow {y}''=\dfrac{-2{{x}^{4}}-2x{{y}^{3}}}{{{y}^{5}}}\]
 \[\begin{align}
  & \Rightarrow {y}''=\dfrac{-2{{x}^{4}}}{{{y}^{5}}}-\dfrac{2x{{y}^{3}}}{{{y}^{5}}} \\
 & \Rightarrow {y}''=\dfrac{-2{{x}^{4}}}{{{y}^{5}}}-\dfrac{2x}{{{y}^{2}}} \\
\end{align}\]

Therefore, the value of \[{y}''\] by implicit differentiation of \[{{x}^{3}}+{{y}^{3}}=1\] is \[{y}''=\dfrac{-2{{x}^{4}}}{{{y}^{5}}}-\dfrac{2x}{{{y}^{2}}}\].

Note: We must remember to find the inner derivative for the functions present in the equation if in case required. In order to solve these types of problems, we must have enough knowledge about derivatives of basic functions. We should avoid calculation mistakes to get the correct solution.