If \(f(x)=x \cdot \cot ^{-1} x\) then \(f^{\prime}(1)=\) (a) \((\pi / 4)-(1 / 2)\) (b) \((\pi / \overline{4)+(1 / 2)}\) (c) \((\pi / 4)-(1 / 3)\) (d) \((\pi / 4)-1\)

Understanding the product rule is crucial for differentiating functions where two or more functions are multiplied together, which is a common occurrence in calculus problems. If you have two functions of a variable, say, \( u(x) \) and \( v(x) \), their derivative is not simply the product of their individual derivatives. Instead, the product rule tells us that \( (uv)' = u'v + uv' \) where \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \) respectively.

Applying this to an exercise involves carefully differentiating each function separately and then following the product rule to combine these derivatives. For students preparing for JEE Maths, mastering the product rule is necessary since it allows you to tackle a variety of complex calculus problems that involve products of functions.

Inverse trigonometric functions are those functions that reverse the action of the basic trigonometric functions. For example, if \( \text{sin}(x) = y \), then \( \text{sin}^{-1}(y) = x \). The inverse trigonometric functions include \( \text{sin}^{-1}(x) \), \( \text{cos}^{-1}(x) \), \( \text{tan}^{-1}(x) \), and similarly for cotangent, which is relevant to our exercise as \( \text{cot}^{-1}(x) \).

The derivatives of these functions have distinct formulas, which are often used in calculus problems. For instance, the derivative of \( \text{cot}^{-1}(x) \) is \( -1/(x^2+1) \), a fact that students should remember. This knowledge is crucial for students to solve problems efficiently in exams such as the JEE, where time management is key to success.

Calculus problems often involve a mix of conceptual knowledge and algebraic manipulation. When solving these problems, it is imperative to recognize which calculus principles apply, such as derivatives, integrals, the chain rule, etcetera, and then to carry out the appropriate operations. For derivatives of functions, one must be comfortable with utilizing rules like power, product, quotient, and chain rules.

In the example provided, both the concept of inverse trigonometric functions and the product rule are used to compute the derivative. It's exercises like these that build the foundational understanding necessary to solve a variety of calculus problems. Students should practice these problems regularly to build fluency and to approach new calculus challenges with confidence.

Preparing for JEE Maths involves a systematic understanding of various mathematical concepts, including calculus. The Joint Entrance Examination (JEE) is highly competitive, and a strong grasp of calculus is necessary for success. Apart from understanding theories, students should aim to solve an extensive range of problems.

Consistent practice with timed tests, familiarity with the format of the questions, and targeted preparation using previous years' papers and mock tests are effective strategies. Specifically for calculus, students should focus on mastering the computation of derivatives and integrals, optimization problems, and the application of different rules for differentiation and integration. The example problem provided in this article is a good representation of the type of calculus problem a student might encounter on the JEE.