Chapter 3.5, Problem 31E

To determine

To calculate: The second derivative of the function $9{\mathrm{x}}^2+{\mathrm{y}}^2=9$ by implicit differentiation.

Answer to Problem 31E

The second derivative of the function is $\mathrm{y}\text{'}\text{'}=\frac{-81}{{\mathrm{y}}^3}$ .

Explanation of Solution

Given information:

The function $9{\mathrm{x}}^2+{\mathrm{y}}^2=9$ .

Formula used:

The chain rule for differentiation is if f is a function of g then $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left(\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right)\right)=\mathrm{f}\text{'}\left(\mathrm{g}\left(\mathrm{x}\right)\right)\mathrm{g}\text{'}\left(\mathrm{x}\right)$ .

Power rule for differentiation is $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}{\mathrm{x}}^{\mathrm{n}}=\mathrm{n}{\mathrm{x}}^{\mathrm{n}-1}$ .

Quotient rule for differentiation is $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left(\frac{\mathrm{f}}{\mathrm{g}}\right)=\frac{\mathrm{g}\left(\mathrm{x}\right)\mathrm{f}\text{'}\left(\mathrm{x}\right)-\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\text{'}\left(\mathrm{x}\right)}{{\left[\mathrm{g}\left(\mathrm{x}\right)\right]}^2}$ where f and g are functions of x .

Calculation:

Consider the function $9{\mathrm{x}}^2+{\mathrm{y}}^2=9$ .

Differentiate both sides with respect to x ,

  $\begin{array}{c}\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left(9{\mathrm{x}}^2+{\mathrm{y}}^2\right)=\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left(9\right)\\ \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left(9{\mathrm{x}}^2\right)+\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left({\mathrm{y}}^2\right)=0\end{array}$

Recall that power rule for differentiation is $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}{\mathrm{x}}^{\mathrm{n}}=\mathrm{n}{\mathrm{x}}^{\mathrm{n}-1}$ and chain rule for differentiation is if f is a function of g then $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left(\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right)\right)=\mathrm{f}\text{'}\left(\mathrm{g}\left(\mathrm{x}\right)\right)\mathrm{g}\text{'}\left(\mathrm{x}\right)$ .

Apply it. Also observe that y is a function of x,

  $\begin{array}{c}\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left(9{\mathrm{x}}^2+{\mathrm{y}}^2\right)=\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left(9\right)\\ \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left(9{\mathrm{x}}^2\right)+\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left({\mathrm{y}}^2\right)=0\\ 18\mathrm{x}+2\mathrm{y}\mathrm{y}\text{'}=0\end{array}$

Isolate the value of $\mathrm{y}\text{'}$ on left hand side and simplify,

  $\begin{array}{c}18\mathrm{x}+2\mathrm{y}\mathrm{y}\text{'}=0\\ 2\mathrm{y}\mathrm{y}\text{'}=-18\mathrm{x}\\ \mathrm{y}\text{'}=\frac{-18\mathrm{x}}{2\mathrm{y}}\\ \mathrm{y}\text{'}=\frac{-9\mathrm{x}}{\mathrm{y}}\end{array}$

Therefore, the value of first derivative of the function is $\mathrm{y}\text{'}=\frac{-9\mathrm{x}}{\mathrm{y}}$ .

Now, again differentiate the above expression with respect to x . Recall that quotient rule for differentiation is $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left(\frac{\mathrm{f}}{\mathrm{g}}\right)=\frac{\mathrm{g}\left(\mathrm{x}\right)\mathrm{f}\text{'}\left(\mathrm{x}\right)-\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\text{'}\left(\mathrm{x}\right)}{{\left[\mathrm{g}\left(\mathrm{x}\right)\right]}^2}$ where f and g are functions of x .

Apply it. Also observe that y is a function of x,

  $\begin{array}{c}\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left(\mathrm{y}\text{'}\right)=\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left(\frac{-9\mathrm{x}}{\mathrm{y}}\right)\\ \mathrm{y}\text{'}\text{'}=\frac{\left(\mathrm{y}\right)\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left(-9\mathrm{x}\right)-\left(-9\mathrm{x}\right)\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left(\mathrm{y}\right)}{{\left(\mathrm{y}\right)}^2}\\ \mathrm{y}\text{'}\text{'}=\frac{\mathrm{y}\left(-9\right)+9\mathrm{x}\left(\mathrm{y}\text{'}\right)}{{\mathrm{y}}^2}\\ \mathrm{y}\text{'}\text{'}=\frac{-9\mathrm{y}+9\mathrm{x}\mathrm{y}\text{'}}{{\mathrm{y}}^2}\end{array}$

Now, substitute the value $\mathrm{y}\text{'}=\frac{-9\mathrm{x}}{\mathrm{y}}$ in the above expression.

  $\begin{array}{c}\mathrm{y}\text{'}\text{'}=\frac{-9\mathrm{y}+9\mathrm{x}\left(\frac{-9\mathrm{x}}{\mathrm{y}}\right)}{{\mathrm{y}}^2}\\ =\frac{-9{\mathrm{y}}^2-81{\mathrm{x}}^2}{{\mathrm{y}}^3}\\ =\frac{-9\left(9{\mathrm{x}}^2+{\mathrm{y}}^2\right)}{{\mathrm{y}}^3}\end{array}$

According to the question $9{\mathrm{x}}^2+{\mathrm{y}}^2=9$ , substitute the value in the above expression,

  $\begin{array}{c}\mathrm{y}\text{'}\text{'}=\frac{-9\left(9{\mathrm{x}}^2+{\mathrm{y}}^2\right)}{{\mathrm{y}}^3}\\ =\frac{-9\left(9\right)}{{\mathrm{y}}^3}\\ =\frac{-81}{{\mathrm{y}}^3}\end{array}$

Thus, the second derivative of the function is $\mathrm{y}\text{'}\text{'}=\frac{-81}{{\mathrm{y}}^3}$ .