Question

If \(\mathrm{y}=\tan ^{-1}\left[1 /\left(\mathrm{x}^{2}+\mathrm{x}+1\right)\right] \tan ^{-1}\left[1 /\left(\mathrm{x}^{2}+3 \mathrm{x}+3\right)\right]\) \(+\tan ^{-1}\left[1 /\left(\mathrm{x}^{2}+5 \mathrm{x}+7\right)\right] \ldots \ldots\) to \(\mathrm{n}\) terms then \((\mathrm{dy} / \mathrm{d} \mathrm{x})\) \(=\) (a) \(\left[1 /\left\\{1+(\mathrm{x}+\mathrm{n})^{2}\right\\}\right]+\left[1 /\left(1+\mathrm{x}^{2}\right)\right]\) (b) \(\left[1 /\left\\{1+(\mathrm{x}+\mathrm{n})^{2}\right\\}\right]-\left[1 /\left(1+\mathrm{x}^{2}\right)\right]\) (c) \(\left[2 /\left\\{1+(\mathrm{x}+\mathrm{n})^{2}\right\\}\right]+\left[1 /\left(1+\mathrm{x}^{2}\right)\right]\) (d) \(\sum \mathrm{n}\)

Short answer

The derivative of the given function with respect to x is: \(\frac{dy}{dx} = \sum_{n=1}^n \left( \frac{(2x+2n-1)(1 + g(x)^2)}{(1 + f(x)^2)(1 + g(x)^2)}\tan^{-1}g(x) + \frac{(2x+2n+1)(1 + f(x)^2)}{(1 + f(x)^2)(1 + g(x)^2)}\tan^{-1}f(x)\right)\) None of the given options match this expression.

Step-by-step solution

Identify each term in the series

For the given function y, each term in the series has the form: \(\tan^{-1}\left[ \frac{1}{(x^2 + (2n - 1)x + 2n^2 - n)}\right] \tan^{-1}\left[ \frac{1}{(x^2 + (2n + 1)x + 2n^2 + n)}\right]\) Where n starts at 1 and goes up to n terms.

Apply the sum rule for derivatives

Knowing that the derivative of a sum is the sum of derivatives, we want to find the derivative of each term in the sum separately and then put them together. For each term, we will apply the product rule and the chain rule to find the derivatives.

Apply the product rule and the chain rule

Now, let's look at an arbitrary term of the series and apply the product rule: \(\frac{d}{dx}\left(\tan^{-1}f(x) \tan^{-1}g(x)\right) = \frac{1}{1 + (f(x))^2}\cdot\frac{df(x)}{dx}\tan^{-1}g(x) + \frac{1}{1 + (g(x))^2}\cdot\frac{dg(x)}{dx}\tan^{-1}f(x)\) This will be our general form for the derivative of each term in the series. We will now apply the chain rule to calculate the derivatives \(\frac{d}{dx}f(x)\) and \(\frac{d}{dx}g(x)\).

Calculate the derivatives df(x)/dx and dg(x)/dx

In order to find \(\frac{d}{dy}f(x)\) and \(\frac{d}{dx}g(x)\), we will apply the chain rule to the fractions \(f(x)\) and \(g(x)\) in our general form above. For example, for f(x): \(\frac{d}{dx}\left(\frac{1}{1+x^2\;+\;(2n-1)x\;+\;2n^2-n}\right) = \frac{d}{dx} \left[ (1+x^2\;+\;(2n-1)x\;+\;2n^2-n)^{-1}\right] = -(1+x^2\;+\;(2n-1)x\;+\;2n^2-n)^{-2}(2x+2n-1)\); and similarly for g(x): \(\frac{d}{dx}\left(\frac{1}{1+x^2\;+\;(2n+1)x\;+\;2n^2+n}\right) = \frac{d}{dx} \left[ (1+x^2\;+\;(2n+1)x\;+\;2n^2+n)^{-1}\right] = -(1+x^2\;+\;(2n+1)x\;+\;2n^2+n)^{-2}(2x+2n+1)\).

Substitute the derivatives into the general form

Now that we have the derivatives for f(x) and g(x), we can substitute these back into our general form for the derivative of each term: \(\frac{d}{dx}\left(\tan^{-1}f(x) \tan^{-1}g(x)\right) = \frac{(2x+2n-1)(1 + g(x)^2)}{(1 + f(x)^2)(1 + g(x)^2)}\tan^{-1}g(x) + \frac{(2x+2n+1)(1 + f(x)^2)}{(1 + f(x)^2)(1 + g(x)^2)}\tan^{-1}f(x)\)

Sum the derivatives of each term in the series

Finally, we will sum the derivatives of each term in the series, applying the sum rule for derivatives: \(\frac{dy}{dx} = \sum_{n=1}^n \left(\frac{d}{dx}\left(\tan^{-1}f(x) \tan^{-1}g(x)\right)\right) = \sum_{n=1}^n \left( \frac{(2x+2n-1)(1 + g(x)^2)}{(1 + f(x)^2)(1 + g(x)^2)}\tan^{-1}g(x) + \frac{(2x+2n+1)(1 + f(x)^2)}{(1 + f(x)^2)(1 + g(x)^2)}\tan^{-1}f(x)\right)\) Which is the derivative of the given function with respect to x.

Compare the derivative to the given options

After examining each option (a, b, c, and d), we can see that none of them matches the derived expression above for the derivative, indicating that the provided options are not correct. To conclude, the derivative of the given function with respect to x does not match any of the given options.

Key concepts

Product Rule

In calculus, the **product rule** is a crucial tool when dealing with derivatives of products of functions. When you have two functions of x, say \(f(x)\) and \(g(x)\), the product rule helps you find the derivative of their product \(f(x) \, g(x)\). The formula for the product rule is:

• \( \frac{d}{dx}[f(x) \, g(x)] = f'(x) \, g(x) + f(x) \, g'(x) \)

This means you differentiate \(f(x)\) while leaving \(g(x)\) unchanged, add to it the differentiation of \(g(x)\) while keeping \(f(x)\) unchanged. In our exercise, we apply this rule to two inverse trigonometric functions multiplied together, i.e., \(\tan^{-1}(f(x)) \tan^{-1}(g(x))\), making it beneficial in computing their derivative efficiently.
This foundational principle makes it easier to understand and break down complex expressions into manageable calculations.

Chain Rule

The **chain rule** is a powerful technique in calculus used to differentiate composite functions. When you have a function inside another function, the chain rule is your go-to method. Say you have a function \(h(x) = f(g(x))\). The chain rule tells us how to find \(h'(x)\):

• \( h'(x) = f'(g(x)) \cdot g'(x) \)

This rule asserts that you differentiate the outer function, leaving the inner function untouched and then multiply it by the derivative of the inner function.
In our exercise, this rule comes into play when we differentiate expressions like \(\tan^{-1}(f(x))\). Here, \(f(x)\) is the function inside in the composition, requiring application of the chain rule to effectively find the derivative, \( \frac{d}{dx}f(x) \) and \( \frac{d}{dx}g(x) \).
Mastering the chain rule is essential for tackling derivatives in more complex configurations where functions are nested inside each other.

Inverse Trigonometric Functions

Understanding **inverse trigonometric functions** is critical in calculus, especially when differentiating functions that include them. Functions like \( \tan^{-1}(x)\) are inverse operations to trigonometric functions and are often used when shifting perspectives from an angle to a ratio or vice versa. The important properties that we use in calculus include their derivatives:

• For \(\tan^{-1}(x)\), the derivative is \(\frac{1}{1 + x^2}\).

These derivatives help us compute complex derivatives involving inverse trigonometric functions.
In our exercise, this comes into play as we repeatedly take derivatives of expressions involving \(\tan^{-1}\), necessitating a clear understanding of these derivatives.
They simplify the complexity of handling inverse trigonometric functions in differentiations, making it a smoother computational process.