Question

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

Answer 1

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 $x'$.