Problem 54 - Calculus - AP Edition

Question

Find \(d y / d x\). $$y=x^{2}\left(\sin ^{-1} x\right)^{3}$$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = 2x(\sin^{-1} x)^3 + \frac{3x^2(\sin^{-1} x)^2}{\sqrt{1-x^2}} \).

Step 1: Identify the Rule to Apply

To find the derivative of the given expression, \( y = x^2 (\sin^{-1} x)^3 \), recognize that it requires the Product Rule since it is a product of two functions: \( u = x^2 \) and \( v = (\sin^{-1} x)^3 \).

Step 2: Differentiate First Function

Differentiate \( u = x^2 \) with respect to \( x \). The derivative \( du/dx = 2x \).

Step 3: Differentiate Second Function Using the Chain Rule

First, rewrite \( v = (\sin^{-1} x)^3 \). Use the Chain Rule for differentiation: \( dv/dx = 3(\sin^{-1} x)^2 \cdot (d/dx)(\sin^{-1} x) \). The derivative of \( \sin^{-1} x \) is \( 1/\sqrt{1-x^2} \). Thus, \( dv/dx = \frac{3(\sin^{-1} x)^2}{\sqrt{1-x^2}} \).

Step 4: Apply the Product Rule

The Product Rule is \( \frac{d}{dx}(uv) = u'v + uv' \). Using the derivatives found, \( \frac{dy}{dx} = (2x)((\sin^{-1} x)^3) + (x^2) \left(\frac{3(\sin^{-1} x)^2}{\sqrt{1-x^2}}\right) \).

Step 5: Simplify the Result

Simplify \( \frac{dy}{dx} = 2x(\sin^{-1} x)^3 + \frac{3x^2(\sin^{-1} x)^2}{\sqrt{1-x^2}} \). There's no further simplification, so this is the result.

Key Concepts

Product Rule

When you're faced with the differentiation of products of functions, the Product Rule really shines. It simplifies the calculus when we deal with two functions being multiplied together, like the expression we're dealing with:

• If you have two functions, say \(u(x)\) and \(v(x)\), then the derivative of their product \(y = u(x) \times v(x)\) is calculated as: \[ \frac{d}{dx}(uv) = u'v + uv' \]
• Essentially, it tells us to differentiate one function, multiply it by the other, then switch. The sum of these gives the derivative of the product.
• In our exercise, \(u = x^2\) and \(v = (\sin^{-1} x)^3\). By applying the Product Rule, we separate the differentiation tasks which lets us handle the problem piece by piece.

Differentiation becomes more straightforward using this rule, especially when functions are simple or complicated like in our case.

Chain Rule

The Chain Rule is another staple of differentiation, particularly useful when dealing with composite functions. When a function is nested inside another, like \(v = (\sin^{-1} x)^3\), employing the Chain Rule is essential. Here's how it works:

• Suppose a function \(g(x)\) is contained within \(f(x)\), i.e., \(y = f(g(x))\). The Chain Rule formula is: \[ \frac{dy}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx} \]
• For our expression, \(v = (\sin^{-1} x)^3\) is really a composite function, where \(g(x) = \sin^{-1} x\), and it further powers \(x\) to three.
• Using the Chain Rule, differentiate the outer function \((3( ext{angle})^2)\) then multiply by the derivative of the inner \( (\sin^{-1} x)\), which is \(\frac{1}{\sqrt{1-x^2}}\).

It's like peeling an onion, taking the derivative layer by layer, ensuring no mistakes in differentiation.

Inverse Trigonometric Functions Differentiation

Inverse trigonometric functions, such as \(\sin^{-1}(x)\), often appear in calculus problems, and understanding their derivatives is crucial. For problems involving these, knowing the differentiation rules at your fingertips speeds up the process:

• The derivative of \(\sin^{-1}(x)\) is particularly important and follows this rule: \[ \frac{d}{dx}(\sin^{-1}(x)) = \frac{1}{\sqrt{1-x^2}} \]
• This stems from the implicit differentiation of sine and its inverse, considering values between \([-1, 1]\).
• Incorporating these derivatives often involves combining them with both the product and chain rules, especially when dealing with expressions like \((\sin^{-1} x)^3\).

By developing a comfort with inverse trigonometric differentiation, tackling complex calculus problems becomes significantly less daunting. Knowing these derivatives by heart aids in seamlessly navigating through differentiation tasks.