Question

If$\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{b}}^{\mathbf{2}}}\mathbf{=}\mathbf{1}$ , find $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}$and$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{2}}}$ by implicit differentiation.

Short answer

The first derivative is $\frac{dy}{dx}=\frac{-b^2}{a^2}\left(\frac{x}{y}\right)$and the second derivative is$\frac{d^{2}y}{d{x}^2}=\frac{-b^4}{a^2{y}^3}$ .

Step-by-step solution

Given Information

The given equation is$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ .

Find the value of dydx

Calculating the first derivative $\frac{dy}{dx}$then we can use it to arrive at a formula for the second derivative $\frac{d^{2}y}{d{x}^2}$using the implicit differentiation.

So, finding the first derivative as:

$$\begin{gathered} \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ \frac{2x}{a^2}+\frac{2y}{b^2}\left(\frac{dy}{dx}\right)=0 \\ \frac{dy}{dx}=\frac{-b^2}{a^2}\left(\frac{x}{y}\right)\left(1\right) \end{gathered}$$

Therefore, the first derivative is$\frac{dy}{dx}=\frac{-b^2}{a^2}\left(\frac{x}{y}\right)$

Find the value of d2ydx2 .

Now finding $\frac{d^{2}y}{d{x}^2}$as:

$\frac{d^{2}y}{d{x}^2}=\frac{-b^2}{a^2}\left(\frac{y-x\frac{dy}{dx}}{y^2}\right)$

Now substituting the value of $\frac{dy}{dx}=\frac{-b^2}{a^2}\left(\frac{x}{y}\right)$as:

$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=\frac{-b^2}{a^2}\left(\frac{y-x\left(\frac{-b^{2}x}{a^{2}y}\right)}{y^2}\right) \\ \frac{d^{2}y}{d{x}^2}=\left(\frac{\frac{-b^2}{a^2}y-\frac{-b^4{x}^5}{a^{4}y}}{y^2}\right) \\ \frac{d^{2}y}{d{x}^2}=\frac{-b^2}{a^2}\left(\frac{y^2+\frac{b^2}{a^2}{x}^2}{y^3}\right) \\ \frac{d^{2}y}{d{x}^2}=\frac{-b^4}{a^2}\left(\frac{\frac{y^2}{b^2}+\frac{x^2}{a^2}}{y^3}\right) \end{gathered}$$

Now substituting $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$in the above equation as:

$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=\frac{-b^4}{a^2}\left(\frac{1}{y^3}\right) \\ \frac{d^{2}y}{d{x}^2}=\frac{-b^4}{a^2{y}^3} \end{gathered}$$

Therefore, the first derivative is $\frac{dy}{dx}=\frac{-b^2}{a^2}\left(\frac{x}{y}\right)$and the second derivative is

$\frac{d^{2}y}{d{x}^2}=\frac{-b^4}{a^2{y}^3}$.