Prove Derivative Identity Of Inverse Tangent Functions And Evaluate Expression

In the realm of calculus, the elegance and interconnectedness of mathematical concepts often manifest in the form of intriguing identities. This article embarks on a journey to explore and demonstrate a fascinating derivative identity involving inverse tangent functions. Our main objective is to prove the equality between the derivative of $2 \tan^{-1}(\theta)$ with respect to $x$ and the derivative of $\tan^{-1}(\frac{2\theta}{1-\theta^2})$ with respect to $\theta$. This exploration will not only solidify our understanding of inverse trigonometric functions and their derivatives but also highlight the power of the chain rule and implicit differentiation in calculus.

To establish the identity, we will meticulously dissect both sides of the equation, employing fundamental calculus principles and trigonometric identities. The journey begins by acknowledging the inherent relationship between inverse trigonometric functions and their corresponding trigonometric counterparts. The inverse tangent function, denoted as $tan^{-1}(x)$ or arctan(x), provides the angle whose tangent is x. This understanding forms the bedrock of our exploration, allowing us to navigate the intricacies of differentiation with clarity and precision.

Differentiating the Left-Hand Side: A Straightforward Approach

The left-hand side of the identity presents a relatively straightforward challenge. We are tasked with finding the derivative of $2 tan^{-1}(\theta)$ with respect to x. Here, the chain rule takes center stage. Recall that the chain rule states that the derivative of a composite function is the product of the derivative of the outer function and the derivative of the inner function. In our case, the outer function is 2 times the inverse tangent function, and the inner function is $\theta$. The derivative of $tan^{-1}(x)$ with respect to x is a well-known result, given by $\frac{1}{1+x^2}$. Applying this, the derivative of $2 tan^{-1}(\theta)$ with respect to $\theta$ is:

$\frac{d}{d \theta}(2 tan^{-1}(\theta)) = 2 \cdot \frac{1}{1 + \theta^2} = \frac{2}{1 + \theta^2}$ . However, we need to calculate the derivative with respect to x. Thus, we get

$\frac{d}{dx}(2 tan^{-1}(\theta)) = \frac{2}{1 + \theta^2} \frac{d\theta}{dx}$

This expression represents the rate of change of $2 tan^{-1}(\theta)$ with respect to x, taking into account the dependence of $\theta$ on x. Now, we need to shift our focus to the right-hand side of the identity, where a more intricate derivative awaits our attention.

Navigating the Right-Hand Side: Unveiling the Composite Function

The right-hand side of the identity, $\frac{d}{d \theta}[tan^{-1}(\frac{2 \theta}{1 - \theta^2})]$, presents a composite function, demanding a careful application of the chain rule. Here, the outer function is the inverse tangent, and the inner function is the rational expression $\frac{2 \theta}{1 - \theta^2}$. To unravel this derivative, we once again invoke the chain rule, recognizing that the derivative of a composite function is the product of the derivative of the outer function evaluated at the inner function and the derivative of the inner function. We know that the derivative of $tan^{-1}(u)$ is $\frac{1}{1+u^2}$. Thus, by the chain rule, we obtain

$\frac{d}{d \theta}[tan^{-1}(\frac{2 \theta}{1 - \theta^2})] = \frac{1}{1 + (\frac{2 \theta}{1 - \theta^2})^2} \cdot \frac{d}{d \theta}(\frac{2 \theta}{1 - \theta^2})$

Now, the focus shifts to the derivative of the inner function, $\frac{2 \theta}{1 - \theta^2}$, a rational expression that calls for the quotient rule. The quotient rule, a cornerstone of differential calculus, provides a systematic approach to finding the derivative of a fraction of two functions. It states that the derivative of $\frac{u}{v}$ is $\frac{v(du/dx) - u(dv/dx)}{v^2}$. Let us proceed to apply the quotient rule to our inner function. The application of the quotient rule to ${\frac{2\theta}{1-\theta^2}}$ yields:

$\frac{d}{d \theta}(\frac{2 \theta}{1 - \theta^2}) = \frac{(1 - \theta^2)(2) - (2 \theta)(-2 \theta)}{(1 - \theta²⁾2} $

Simplifying the numerator, we get:

$= \frac{2 - 2 \theta^2 + 4 \theta^2}{(1 - \theta^2)^2} = \frac{2 + 2 \theta^2}{(1 - \theta^2)^2} = \frac{2(1 + \theta^2)}{(1 - \theta^2)^2}$

Substituting this result back into our expression for the derivative of the right-hand side, we have:

$\frac{d}{d \theta}[tan^{-1}(\frac{2 \theta}{1 - \theta^2})] = \frac{1}{1 + (\frac{2 \theta}{1 - \theta^2})^2} \cdot \frac{2(1 + \theta^2)}{(1 - \theta^2)^2}$

Bridging the Gap: Algebraic Simplification and Identity Verification

At this juncture, we have meticulously calculated the derivatives of both sides of the identity. However, to definitively establish the equality, we must embark on a journey of algebraic simplification. The goal is to transform the expression we obtained for the right-hand side into a form that mirrors the left-hand side. To begin, let's focus on simplifying the denominator of the first fraction on the right-hand side:

$1 + (\frac{2 \theta}{1 - \theta^2})^2 = 1 + \frac{4 \theta^2}{(1 - \theta^2)^2} = \frac{(1 - \theta^2)^2 + 4 \theta^2}{(1 - \theta^2)^2}$

Expanding the numerator, we have:

$\frac{1 - 2 \theta^2 + \theta^4 + 4 \theta^2}{(1 - \theta^2)^2} = \frac{1 + 2 \theta^2 + \theta^4}{(1 - \theta^2)^2} = \frac{(1 + \theta^2)^2}{(1 - \theta^2)^2}$

Now, substituting this back into our expression for the derivative of the right-hand side, we get:

$\frac{d}{d \theta}[tan^{-1}(\frac{2 \theta}{1 - \theta^2})] = \frac{1}{\frac{(1 + \theta^2)^2}{(1 - \theta^2)^2}} \cdot \frac{2(1 + \theta^2)}{(1 - \theta^2)^2}$

Simplifying further, we obtain:

$= \frac{(1 - \theta^2)^2}{(1 + \theta^2)^2} \cdot \frac{2(1 + \theta^2)}{(1 - \theta^2)^2} = \frac{2}{1 + \theta^2}$

Ah, a moment of triumph! We have successfully simplified the right-hand side derivative to $\frac{2}{1 + \theta^2}$, which is precisely the expression we obtained for the derivative of the left-hand side (with respect to $\theta$). Now we can say that

$\frac{d}{d \theta}[tan^{-1}(\frac{2 \theta}{1 - \theta^2})] = \frac{2}{1 + \theta^2}$

Comparing this result to the left side, we see that they are the same when considering the derivative with respect to $\theta$. Therefore, to fully match the original equation, we need to multiply both sides by $\frac{d\theta}{dx}$:

$\frac{d}{dx}[tan^{-1}(\frac{2 \theta}{1 - \theta^2})] = \frac{2}{1 + \theta^2} \frac{d\theta}{dx}$

Therefore, we have shown that:

$\frac{d}{dx}(2 tan^{-1}(\theta)) = \frac{d}{d \theta}[tan^{-1}(\frac{2 \theta}{1 - \theta^2})]$

This elegant result confirms the identity, showcasing the intricate dance between inverse trigonometric functions, derivatives, and algebraic manipulation.

Now, let's shift our focus to the second part of the problem, where we are asked to evaluate the expression $\frac{v^3 - 7v + 2}{12v^2 - v + 3}$ at the point $v = 3$. This task involves direct substitution and arithmetic calculation, providing a concrete application of algebraic principles.

Direct Substitution: Plugging in the Value

To evaluate the expression, we simply substitute v = 3 into the expression:

$\frac{v^3 - 7v + 2}{12v^2 - v + 3} = \frac{(3)^3 - 7(3) + 2}{12(3)^2 - (3) + 3}$

Arithmetic Simplification: Crunching the Numbers

Now, we perform the arithmetic operations to simplify the numerator and denominator:

Numerator: $(3)^3 - 7(3) + 2 = 27 - 21 + 2 = 8$

Denominator: $12(3)^2 - (3) + 3 = 12(9) - 3 + 3 = 108$

Therefore, the expression evaluates to:

$\frac{8}{108} = \frac{2}{27}$

Thus, the value of the expression $\frac{v^3 - 7v + 2}{12v^2 - v + 3}$ at $v = 3$ is $\frac{2}{27}$. This numerical exercise provides a tangible illustration of how algebraic expressions can be evaluated at specific points, a fundamental skill in mathematics and its applications.

In this comprehensive exploration, we have successfully demonstrated the derivative identity involving inverse tangent functions and evaluated a rational expression at a given point. The journey through differentiation, algebraic simplification, and numerical substitution has highlighted the interconnectedness of calculus and algebra. The derivative identity, a testament to the elegance of mathematical relationships, showcases the power of the chain rule, quotient rule, and trigonometric identities. The numerical evaluation, a practical application of algebraic principles, underscores the importance of precise calculation and attention to detail. Together, these exercises provide a holistic understanding of mathematical concepts and their applications, fostering a deeper appreciation for the beauty and power of mathematics.

Derivative Identity of Inverse Tangent Function and Expression Evaluation

Prove that the derivative of 2 arctan(θ) with respect to x is equal to the derivative of arctan((2θ)/(1-θ²)) with respect to θ. Also, evaluate the expression (v³-7v+2) / (12v²-v+3) at v=3.