Problem 294 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { If } y=x^{2} \sin x, \text { find } \frac{d^{25} y}{d x^{25}} \text { . } $$

Step-by-Step Solution

Answer

The 25th derivative of the function \(y = x^2 \sin(x)\) will have the form: \( \frac{d^{25} y}{dx^{25}} = A \sin(x) - Bx^{22} \cos(x) \) with A and B being some coefficients. To find the exact values of A and B, an iterative process or a computer algorithm can be employed.

Step 1: Finding the first derivative

We start by finding the first derivative of the function. Using the product rule for differentiation, which is (uv)' = u'v + uv': \( y = x^2 \sin(x) \) \( \frac{dy}{dx} = (x^2)' \sin(x) + x^2 (\sin(x))' \) \( \frac{dy}{dx} = 2x \sin(x) + x^2 \cos(x) \)

Step 2: Finding the second derivative

Now we find the second derivative by differentiating the first derivative with respect to x: \( \frac{d^2 y}{dx^2} = \frac{d}{dx} (2x \sin(x) + x^2 \cos(x)) \) Using the product rule again: \( \frac{d^2 y}{dx^2} = (2x)' \sin(x) + 2x (\sin(x))' + (x^2)' \cos(x) + x^2 (\cos(x))' \) \( \frac{d^2 y}{dx^2} = 2 \sin(x) + 2x \cos(x) - 2x \cos(x) - x^2 \sin(x) \) Simplifying: \( \frac{d^2 y}{dx^2} = 2 \sin(x) - x^2 \sin(x) \)

Step 3: Observing the pattern

Now let's look at the first and second derivatives we have found: First derivative: \( \frac{dy}{dx} = 2x \sin(x) + x^2 \cos(x) \) Second derivative: \( \frac{d^2 y}{dx^2} = 2 \sin(x) - x^2 \sin(x) \) We observe that every derivative has two terms, one with a sine function and one with its cosine counterpart. The sine term always has a negative sign, and it seems that the power of x is reduced by 2 in each derivative. We can assume that the pattern continues and that in every odd derivative (1st, 3rd, 5th, etc.) the cosine term will have an even power of x and the sine term will have an odd power of x. Conversely, every even derivative will have even powers of x in the sine term and odd powers of x in the cosine term. We are interested in the 25th derivative, which is an odd-order derivative, so the sine term will have an odd power of x while the cosine term will have an even power of x.

Step 4: Finding the 25th derivative

Based on our assumption of the pattern so far, the 25th derivative will have the general form: \( \frac{d^{25} y}{dx^{25}} = A \sin(x) - Bx \cos(x) \) To find the coefficients A and B, we need to go through the process of finding the derivatives until we reach the 25th derivative. The pattern, however, already provides us with the necessary information. In the first derivative, the power of x in the sine term is 1, and in the cosine term, it is 2. In the second derivative, the power of x in the sine term is 0, and in the cosine term, it is 2. With this observation, we conclude that, in the 25th derivative, the power of x in the sine term would be the nearest lesser odd integer to 25, which is 23 (since it's an odd-order derivative). Similarly, the cosine term will have a power of x that is even and one less than the power of x in the sine term, which is 22. Thus, the 25th derivative will have the form: \( \frac{d^{25} y}{dx^{25}} = A \sin(x) - Bx^{22} \cos(x) \) Calculating the coefficients A and B is quite challenging, and it might be easier to approach this through a computer algorithm that iteratively finds the coefficients of A and B.

Key Concepts

Product Rule

In differential calculus, the product rule is a fundamental technique used for differentiating products of two functions. When you have a function that is the product of two simpler functions, say \(u(x)\) and \(v(x)\), the product rule tells you how to differentiate this composite function. The rule states:

• If \(y = u(x) \cdot v(x)\), then \(\frac{dy}{dx} = u'(x) \cdot v(x) + u(x) \cdot v'(x)\).

This simple formula allows us to break down the differentiation of the product into manageable parts. First, you differentiate \(u(x)\) leaving \(v(x)\) unchanged, and then you add the product where \(v(x)\) is differentiated, keeping \(u(x)\) as it is. By applying the product rule consistently, we can tackle more complex differentiation problems easily.
In the example exercise, \(x^2\) is \(u(x)\) and \(\sin(x)\) is \(v(x)\), leading to the first derivative using the product rule: \(2x \sin(x) + x^2 \cos(x)\).

Higher-Order Derivatives

Higher-order derivatives refer to the process of differentiating a function multiple times. The first derivative of a function gives us the rate of change, or slope, of the function. Higher-order derivatives, such as the second, third, or even the 25th as in this exercise, provide deeper insights into the behavior of the function.
When taking higher-order derivatives, patterns often emerge. For example, the sign and degree of coefficients can repeat at regular intervals. In our original problem, we noted a distinct pattern in how the powers of \(x\) were decreasing by 2 with each derivative. This kind of pattern helps streamline the process of finding very high order derivatives without manual calculation of each step.
These patterns can also indicate the type of function involved, such as trigonometric or polynomial functions, and reveal symmetries or periodicities in the function's behavior. Recognizing and utilizing these patterns is crucial for efficiently calculating higher derivatives without extensive computation.

Trigonometric Functions

Trigonometric functions like \(\sin\) and \(\cos\) are fundamental to differential calculus, especially when integrated with polynomial terms. They oscillate between values and introduce interesting behavior into derivatives. The derivatives of these functions have tidy results:

• The derivative of \(\sin(x)\) is \(\cos(x)\).
• The derivative of \(\cos(x)\) is \(-\sin(x)\).

In calculations, these transformations can sometimes produce alternating signs and periodicity to the derivative's terms. For example, in our problem, every derivative involves sine and cosine terms where the roles of \(\sin(x)\) and \(\cos(x)\) switch based on whether it is an even or odd derivative.
The consistent pattern derived from the trigonometric derivatives helps in predicting the nature of higher derivatives, as we did with the 25th derivative in the exercise. Understanding how these functions work and change with differentiation is crucial to mastering problems involving higher-order derivatives.