Question

Find \(\frac{d y}{d x}\) using implicit differentiation. $$x^{2} \tan y=50$$

Short answer

The derivative is \(\frac{dy}{dx} = \frac{-2 \sin y \cos y}{x}\).

Step-by-step solution

Differentiate with Respect to x

To find \(\frac{dy}{dx}\), we will differentiate both sides of the equation \(x^2 \tan y = 50\) with respect to \(x\). Start by applying the product rule to the left side since it is a product of \(x^2\) and \(\tan y\). The product rule is \(\frac{d}{dx}[uv] = u\frac{dv}{dx} + v\frac{du}{dx}\).

Apply Product Rule

Let \(u = x^2\) and \(v = \tan y\). Differentiate \(u\) with respect to \(x\) to get \(\frac{du}{dx} = 2x\). Now differentiate \(v = \tan y\) with respect to \(x\), using the chain rule: \(\frac{dv}{dx} = \sec^2 y \cdot \frac{dy}{dx}\).

Substitute Derivatives Back

Substitute the derivatives back into the product rule formula: \(x^2 \cdot \sec^2 y \cdot \frac{dy}{dx} + \tan y \cdot 2x = 0\).

Solve for \(\frac{dy}{dx}\)

Rearrange the equation to solve for \(\frac{dy}{dx}\): \(x^2 \sec^2 y \frac{dy}{dx} = -2x \tan y\).Now, divide both sides by \(x^2 \sec^2 y\): \(\frac{dy}{dx} = \frac{-2x \tan y}{x^2 \sec^2 y}\).

Simplify the Expression

Simplify the expression for \(\frac{dy}{dx}\) by canceling common terms: \(\frac{dy}{dx} = \frac{-2 \tan y}{x \sec^2 y}\). Since \(\sec^2 y = \frac{1}{\cos^2 y}\), rewrite \(\tan y\) and \(\sec^2 y\) to get:\(\frac{dy}{dx} = \frac{-2 \sin y \cos y}{x}\).

Key concepts

Understanding the Product Rule

The product rule is an essential concept in calculus when dealing with derivatives of products of functions. In our problem, we have the equation \(x^2 \tan y = 50\). Here, the left side consists of two multiplicative components, \(x^2\) and \(\tan y\), indicating that the product rule is needed. Taking the derivative of a product requires you to separately find the derivative of each component and then combine them as follows:

• Consider \(u = x^2\) and \(v = \tan y\).
• The derivative of a product \(uv\) is given by \(u\frac{dv}{dx} + v\frac{du}{dx}\).
• For our example, \(\frac{du}{dx} = 2x\) since the derivative of \(x^2\) with respect to \(x\) is straightforward.

This step sets the foundation for correctly approaching the derivative calculation of more complex expressions that involve products of two functions.

Diving Into the Chain Rule

When differentiating complex trigonometric functions such as \(\tan y\) with respect to \(x\), the chain rule plays a crucial role. The chain rule helps us differentiate compositions of functions, a situation common in implicit differentiation.Here's how the chain rule is applied to \(\tan y\):

• Let's take \(v = \tan y\). We can't directly differentiate with respect to \(x\) as \(y\) also needs differentiation.
• Using the chain rule, differentiate \(\tan y\) to get \(\frac{dv}{dx} = \sec^2 y \cdot \frac{dy}{dx}\).This uses the known derivative of \(\tan y\), which is \(\sec^2 y\), multiplied by \(\frac{dy}{dx}\), capturing the fact that \(y\) relies on \(x\).

The use of the chain rule here allows us to correctly compute derivatives where one function is nested within another. This ensures that all variables dependent on \(x\) are accurately accounted for.

Trigonometric Differentiation Essentials

In our given exercise, trigonometric differentiation is vital since we are dealing with \(\tan y\). Trigonometric functions, such as sine, cosine, and tangent, have standard derivatives that one must memorize to efficiently tackle differential calculus problems.Key points for differentiating trigonometric functions include:

• The derivative of \(\tan y\) is \(\sec^2 y\), a fact used when applying the chain rule.
• Additionally, the identity \(\sec^2 y = 1/\cos^2 y\) is useful when simplifying expressions.
• In the solution, this relationship assists in rewriting the final expression for \(\frac{dy}{dx}\) in terms of sine and cosine through the identities \(\tan y = \sin y/\cos y\) and \(\sec y = 1/\cos y\).

Understanding these derivatives and trigonometric identities is crucial for properly working through derivatives that involve trigonometric functions, ultimately leading to simplified and correct solutions in implicit differentiation scenarios.