Problem 47 - Calculus - AP Edition

Question

Find \(d y / d x\). $$y=\tan ^{-1}\left(x^{3}\right)$$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = \frac{3x^2}{1+x^6}\)

Step 1: Identify the Function to Differentiate

We need to differentiate the function \(y = \tan^{-1}(x^3)\) with respect to \(x\). The function consists of the inverse tangent of \(x^3\).

Step 2: Apply the Chain Rule

To differentiate \(tan^{-1}(x^3)\), we'll use the chain rule: \(\frac{dy}{dx} = \frac{d}{dx} \tan^{-1}(u)\cdot \frac{du}{dx}\), where \(u = x^3\).

Step 3: Differentiate the Outer Function

Differentiate the outer function \(\tan^{-1}(u)\). The derivative of \(\tan^{-1}(u)\) with respect to \(u\) is \(\frac{1}{1+u^2}\).

Step 4: Differentiate the Inner Function

Differentiate the inner function \(u = x^3\) with respect to \(x\). The derivative is \(\frac{du}{dx} = 3x^2\).

Step 5: Combine Using the Chain Rule

Apply the chain rule: \(\frac{dy}{dx} = \frac{1}{1+(x^3)^2} \cdot 3x^2 = \frac{3x^2}{1+x^6}\). This is the derivative of \(y\) with respect to \(x\).

Key Concepts

Chain Rule

The chain rule is a vital tool in calculus that helps simplify the differentiation of composite functions.
The essence of the chain rule is to break down complex calculations into simpler parts that are more manageable.

• The chain rule states that the derivative of a composite function, like \(f(g(x))\), can be calculated as the derivative of the outer function with respect to the inner function, multiplied by the derivative of the inner function with respect to the original variable. In formula terms, this is \((f(g(x)))'=f'(g(x)) \cdot g'(x)\).
• To apply it, first identify the inner and outer functions. In our example, \(y = \tan^{-1}(x^3)\), the outer function is \(\tan^{-1}(u)\) where \(u = x^3\).
• We then separately find the derivatives: the derivative of \(\tan^{-1}(u)\) with respect to \(u\) and the derivative of \(x^3\) with respect to \(x\).

Finally, both derivatives are multiplied together to find the overall derivative of the function. This way, the chain rule helps us navigate through layers of functions with ease.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as \(\tan^{-1}(x)\), are the inverses of the basic trigonometric functions.
They are quite useful when we need to determine angles from given trigonometric values. They range within specific intervals to ensure that each function remains a proper inverse.

• For instance, \(\tan^{-1}(x)\) is the inverse of the tangent function, which accepts any real number and returns an angle between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).
• These functions have special derivatives. For \(\tan^{-1}(x)\), its derivative is \(\frac{1}{1+x^2}\).
• When differentiating expressions with these inverse functions, we often use the chain rule to handle any nested components like in our example where \(x^3\) is nested inside \(\tan^{-1}\).

Understandably, recognizing and applying these derivatives efficiently requires practice, but once mastered, they are powerful tools in answering many calculus problems.

Calculus

Calculus, as a field, centers on two primary concepts: differentiation and integration.
In solving the type of problem we see here, differentiation takes the spotlight.

• Differentiation involves finding how a function changes at any point, essentially determining its rate of change or slope at specific points.
• The derivative function provides us with a new function that describes this rate of change.
• In our problem, differentiating \(y = \tan^{-1}(x^3)\) means we explored how the angle \(y\) changes with varying \(x\).
• It's worth noting the powerful utility provided by calculus in various fields like physics, engineering, and economics, where analyzing change is fundamental.

By understanding calculus concepts and methods such as the chain rule and the properties of inverse trigonometric functions, you'll be well-equipped to tackle complex mathematical challenges.