Problem 1 - Applied Calculus

Question

(a) Find the derivatives of \(\sin \left(x^{2}+1\right)\) and \(\sin \left(x^{3}+1\right)\) (b) Use your answer to part (a) to find antiderivatives of: (i) \(x \cos \left(x^{2}+1\right)\) (ii) \(x^{2} \cos \left(x^{3}+1\right)\) (c) Find the general antiderivatives of: (i) \(x \sin \left(x^{2}+1\right)\) (ii) \(x^{2} \sin \left(x^{3}+1\right)\)

Step-by-Step Solution

Answer

(a) 2x\cos(x^2+1), 3x^2\cos(x^3+1); (b) \frac{1}{2}\sin(x^2+1)+C, \frac{1}{3}\sin(x^3+1)+C; (c) Not directly solvable, requires advanced techniques.

Step 1: Derivative of \( \sin(x^2 + 1) \)

To find the derivative of \( \sin(x^2 + 1) \), we use the chain rule. Let \( u = x^2 + 1 \). Then, the derivative \( \frac{d}{dx} \sin(u) = \cos(u) \frac{du}{dx} \). Since \( u = x^2 + 1 \), \( \frac{du}{dx} = 2x \). Hence, \( \frac{d}{dx} \sin(x^2 + 1) = \cos(x^2 + 1) \cdot 2x = 2x \cos(x^2 + 1) \).

Step 2: Derivative of \( \sin(x^3 + 1) \)

Using the same approach, let \( u = x^3 + 1 \). The derivative \( \frac{d}{dx} \sin(u) = \cos(u) \frac{du}{dx} \). Here \( \frac{du}{dx} = 3x^2 \) because \( u = x^3 + 1 \). Therefore, \( \frac{d}{dx} \sin(x^3 + 1) = \cos(x^3 + 1) \cdot 3x^2 = 3x^2 \cos(x^3 + 1) \).

Step 3: Antiderivative of \( x \cos(x^2 + 1) \)

From Step 1, \( \frac{d}{dx} \sin(x^2 + 1) = 2x \cos(x^2 + 1) \). Thus, the antiderivative of \( x \cos(x^2 + 1) \) is \( \int x \cos(x^2 + 1) \, dx = \frac{1}{2} \sin(x^2 + 1) + C \).

Step 4: Antiderivative of \( x^2 \cos(x^3 + 1) \)

From Step 2, \( \frac{d}{dx} \sin(x^3 + 1) = 3x^2 \cos(x^3 + 1) \). Therefore, \( \int x^2 \cos(x^3 + 1) \, dx = \frac{1}{3} \sin(x^3 + 1) + C \).

Step 5: General antiderivative of \( x \sin(x^2 + 1) \)

Using integration by parts where \( u = x \) and \( dv = \sin(x^2 + 1) dx \), we find \( v \) involves a substitution based on \( x \cos(x^2 + 1) \). Calculating this, assuming direct integration isn't straightforward, use substitution or parts to obtain the result, which after all steps could typically result in an expression involving polylogarithms or numerical methods. Generally, solving from primitives is required.

Step 6: General antiderivative of \( x^2 \sin(x^3 + 1) \)

Similar to Step 5, use integration by parts, \( u = x^2 \), \( dv = \sin(x^3 + 1) dx \), attempt integration by checking possible solutions, evaluating simpler expressions, or using numerical or series expansions if direct integration is complex. This step requires methods that simplify or handle difficult antiderivatives numerically or with special functions.

Key Concepts

Derivatives

In calculus, derivatives measure how a function changes as its input changes. When you see a function like \( \sin(x^2 + 1) \), you're looking at a composite function, meaning it's a function (\( \sin(x) \)) within another function (\( x^2 + 1 \)). To find the derivative of a composite function, we employ a powerful method known as the Chain Rule.

Here’s how it works:

• Identify the inner function (often represented as \( u \)), which in this case is \( x^2 + 1 \).
• Find the derivative of the outer function \( \sin(u) \) with respect to \( u \), giving us \( \cos(u) \).
• Multiply this by the derivative of the inner function \( \frac{du}{dx} = 2x \) for \( \sin(x^2 + 1) \) or \( 3x^2 \) for \( \sin(x^3 + 1) \).

The final derivatives become \( 2x \cos(x^2 + 1) \) for \( \sin(x^2 + 1) \) and \( 3x^2 \cos(x^3 + 1) \) for \( \sin(x^3 + 1) \). These derivatives tell us how sensitive the original sine functions are to changes in \( x \).

Antiderivatives

Antiderivatives, in essence, work in reverse of derivatives. While derivatives find how a function changes, antiderivatives are about finding the original function from its rate of change. For example, given \( x \cos(x^2 + 1) \), we already derived that \( \frac{d}{dx} \sin(x^2 + 1) = 2x \cos(x^2 + 1) \). Hence, to find its antiderivative, utilize this relationship.

The key is:

• The antiderivative of \( x \cos(x^2 + 1) \) is half of what we differentiate in Step 1’s derivative, which gives \( \frac{1}{2} \sin(x^2 + 1) + C \).
• For \( x^2 \cos(x^3 + 1) \), use a similar method: knowing the derivative from Step 2, the solution is \( \frac{1}{3} \sin(x^3 + 1) + C \).

Here, \( C \) is the constant of integration, accounting for any constant that could be added to our original function, as it disappears during differentiation.

Chain Rule

The Chain Rule is a fundamental tool in calculus for differentiating composite functions. This rule breaks down the differentiation of complex functions into simpler parts. If you have a composite function \( f(g(x)) \), the Chain Rule states that the derivative is \( f'(g(x)) \cdot g'(x) \).

Applying the Chain Rule to \( \sin(x^2 + 1) \) or \( \sin(x^3 + 1) \), we treated \( x^2 + 1 \) and \( x^3 + 1 \) as the inner functions:

• First, differentiate the outer function \( \sin(u) \), resulting in \( \cos(u) \).
• Then multiply by the derivative of the inner parts, which were \( 2x \) and \( 3x^2 \) respectively.

This principle simplifies finding derivatives of even very complex compositions, by essentially unwinding them step by step and applying simple derivative rules to the components.