Question

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

Short answer

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

Step-by-step solution

Differentiate both sides with respect to x

Start by differentiating both sides of the given equation with respect to x. Applying the implicit differentiation to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we get $\frac{d}{dx}\frac{x^2}{a^2}+\frac{d}{dx}\frac{y^2}{b^2}=\frac{d}{dx}(1)$.

Simplify the first term

Differentiate the first term $\frac{x^2}{a^2}$ with respect to x: $\frac{d}{dx}\frac{x^2}{a^2}=\frac{2x}{a^2}$.

Apply chain rule to the second term

Using the chain rule for the second term $\frac{y^2}{b^2}$, we get $\frac{d}{dx}\frac{y^2}{b^2}=\frac{2y}{b^2}\frac{dy}{dx}$.

Combine the results

Combining these results, the differentiation gives: $\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0$.

Solve for $\frac{dy}{dx}$

Rearrange the equation to solve for $\frac{dy}{dx}$: $\frac{2y}{b^2}\frac{dy}{dx}=-\frac{2x}{a^2}$, hence $\frac{dy}{dx}=-\frac{x{a}^2}{y{b}^2}$.

Differentiate $\frac{dy}{dx}$ implicitly

Next, differentiate $\frac{dy}{dx}=-\frac{x{a}^2}{y{b}^2}$ implicitly again with respect to x: Use the quotient rule, letting \ u = -\frac{xa^2}{yb^2}\. Differentiate each part: $\frac{d}{dx}\frac{dy}{dx}=\frac{d^{2}y}{d{x}^2}$.

Apply the quotient rule

Differentiate using quotient rule: $\frac{d}{dx}\frac{-x{a}^2}{y{b}^2}=\frac{(-a^{2}y{b}^2-(-x{a}^2)\frac{dy}{dx}{b}^2)}{(y{b}^2{)}^2}=\frac{-a^{2}y{b}^2+x{a}^2\frac{x{a}^2}{y{b}^2}{b}^2}{(y{b}^2{)}^2}$.

Simplify the result

Simplify the differentiation result to get $\frac{d^{2}y}{d{x}^2}=\frac{a^2(b^2{y}^3+x^2)}{y^3{b}^4}$.

Key concepts

First Derivative

The first derivative represents the rate at which a function is changing at any given point. To find the first derivative $\frac{dy}{dx}$ of the given equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we use implicit differentiation.

Start by differentiating each term with respect to $x$.

The derivative of $\frac{x^2}{a^2}$ with respect to $x$ is $\frac{2x}{a^2}$.

Then, for the second term $\frac{y^2}{b^2}$, since $y$ is a function of $x$, we apply the chain rule and differentiate $y^2$ with respect to $x$ to get $\frac{2y}{b^2}\frac{dy}{dx}$.

Combining these, we get: $\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0$.

Solving for $\frac{dy}{dx}$, we find: $\frac{dy}{dx}=-\frac{a^{2}x}{b^{2}y}$.

Second Derivative

The second derivative, $\frac{d^{2}y}{d{x}^2}$, represents the rate of change of the first derivative. It gives us information about the concavity and the inflection points of the function. To find the second derivative, we differentiate $\frac{dy}{dx}$ again with respect to $x$.

From the first derivative, $\frac{dy}{dx}=-\frac{a^{2}x}{y{b}^2}$.

To differentiate this, we need to apply the quotient rule.

Let $u=-a^{2}x$ and $v=y{b}^2$.

Applying the quotient rule: $\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$.

Here, $\frac{du}{dx}=-a^2$, and by applying the chain rule, $\frac{dv}{dx}=b^2\frac{dy}{dx}$.

Substitute back, we get: $\frac{d^{2}y}{d{x}^2}=\frac{-a^2{b}^{2}y+a^{2}x\frac{dy}{dx}{b}^2}{(y{b}^2{)}^2}$.

Simplifying further, we get: $\frac{d^{2}y}{d{x}^2}=\frac{a^2(b^2{y}^3+x^2)}{y^3{b}^4}$.

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate composite functions. It states that if a function $y$ depends on $u$, which in turn depends on $x$, then the derivative of $y$ with respect to $x$ is: $\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$.

In our example, to differentiate $\frac{y^2}{b^2}$ with respect to $x$, we treat $y$ as a function of $x$:

First, differentiate $y^2$ with respect to $y$ to get $2y$.

Then, multiply by the derivative of $y$ with respect to $x$: $\frac{dy}{dx}$.

Thus, we get: $\frac{d}{dx}(\frac{y^2}{b^2})=\frac{2y}{b^2}\cdot \frac{dy}{dx}$.

This approach simplifies the process of differentiating composite functions.

Quotient Rule

The quotient rule is used to differentiate functions where one function is divided by another. It states that for functions $u(x)$ and $v(x)$, the derivative of $\frac{u}{v}$ is given by: $\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$.

In our example, we apply the quotient rule to $\frac{-a^{2}x}{y{b}^2}$ to find the second derivative.

Let $u=-a^{2}x$ and $v=y{b}^2$.

We have $\frac{du}{dx}=-a^2$ and using the chain rule, $\frac{dv}{dx}=b^2\cdot \frac{dy}{dx}$.

Substituting these into the quotient rule formula, we get: $\frac{d}{dx}(\frac{-a^{2}x}{y{b}^2})=\frac{y{b}^2\cdot (-a^2)-(-a^{2}x)\cdot \frac{dv}{dx}}{(y{b}^2{)}^2}$.

Simplifying this, we find: $\frac{d^{2}y}{d{x}^2}=\frac{a^2(b^2{y}^3+x^2)}{y^3{b}^4}$.