How do you find \[y''\] by implicit differentiation for \[4{x^2} + 3{y^2} = 6\]?

Hint: Here we have to double differentiate the given equation. we will differentiate both sides of the given equation with respect to \[x\]. Then we will take all the \[\dfrac{{dy}}{{dx}}\] terms on one side and then again differentiate both sides with respect to \[x\]. Finally, we will replace the value of \[\dfrac{{dy}}{{dx}}\] in the equation and solve it to get the required value.

Complete step-by-step answer:
We have to find \[y''\] of
\[4{x^2} + 3{y^2} = 6\]
Now differentiating both the side of the equation with respect to \[x\], we get
\[ \Rightarrow \dfrac{{d\left( {4{x^2}} \right)}}{{dx}} + \dfrac{{d\left( {3{y^2}} \right)}}{{dx}} = \dfrac{{d\left( 6 \right)}}{{dx}}\]
Now using the formula \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\] and \[\dfrac{{d\left( k \right)}}{{dx}} = 0\], we get
\[\begin{array}{l} \Rightarrow 4 \times 2 \times {x^{2 - 1}} + 3 \times 2 \times {y^{2 - 1}}\dfrac{{dy}}{{dx}} = 0\\ \Rightarrow 8x + 6y\dfrac{{dy}}{{dx}} = 0\end{array}\]
Subtracting \[8x\] from both sides, we get
\[ \Rightarrow 6y\dfrac{{dy}}{{dx}} = - 8x\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{8x}}{{6y}} = - \dfrac{{4x}}{{3y}}\]…………………….\[\left( 1 \right)\]
Now, differentiating equation \[\left( 1 \right)\] with respect to \[x\], we get
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{d}{{dx}}\left( {\dfrac{{ - 4x}}{{3y}}} \right)\]
Now using quotient rule \[\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}\] on right side, we get
\[\begin{array}{l} \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{3y \times \dfrac{{d\left( { - 4x} \right)}}{{dx}} - \left( { - 4x} \right)\dfrac{{d\left( {3y} \right)}}{{dx}}}}{{{{\left( {3y} \right)}^2}}}\\ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{3y \times \left( { - 4} \right) - \left( { - 4x} \right) \times 3\dfrac{{dy}}{{dx}}}}{{9{y^2}}}\end{array}\]
Multiplying the terms, we get
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{ - 12y + 12x\dfrac{{dy}}{{dx}}}}{{9{y^2}}}\]
Next, replacing the value of \[\dfrac{{dy}}{{dx}}\] in above equation, we get
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{ - 12y + \left( {12x \times \dfrac{{ - 4x}}{{3y}}} \right)}}{{9{y^2}}}\]
Multiplying the terms, we get
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{ - 12y + \left( {\dfrac{{ - 16{x^2}}}{y}} \right)}}{{9{y^2}}}\]
Taking L.C.M in the numerator, we get
\[\begin{array}{l} \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{\dfrac{{ - 12{y^2} - 16{x^2}}}{y}}}{{9{y^2}}}\\ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{ - 12{y^2} - 16{x^2}}}{{9{y^3}}}\end{array}\]
We can further simplify the above value by taking \[ - 4\] common in numerator as,
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{ - 4\left( {3{y^2} + 4{x^2}} \right)}}{{9{y^3}}}\]
Substituting the value of the given equation \[4{x^2} + 3{y^2} = 6\] in above equation, we get
\[\begin{array}{l} \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{ - 4 \times 6}}{{9{y^3}}}\\ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{ - 8}}{{3{y^3}}}\end{array}\]

So the value of \[y''\] is \[\dfrac{{ - 8}}{{3{y^3}}}\].

Note:
Implicit differentiation is used when we have implicit functions where one function doesn’t lead to another one. Implicit differentiation is done by using the chain rule and viewing \[y\] as an implicit function of \[x\]. There are various case of implicit function which in which we have \[y\] equal to a function that contains \[x,y'\] or \[y'\] in similar way we can have \[x\] equal to a function that contains \[y,x'\] or just \[x'\].