Question

If \(y=\left[\left(\sin ^{2} x\right) /(1-\cot x)\right]+\left[\left(\cos ^{2} x\right) /(1-\tan x)\right]\) and \(\left[\left.[\mathrm{dy} / \mathrm{dx}]\right|_{\mathrm{X}=(\pi / 4)}=\right.\) (a) 0 (b) \(+1\) (c) (d) \(|(1 / 2)|\)

Short answer

The given expression is undefined at \(x = \frac{\pi}{4}\), so none of the options presented are correct due to the indeterminate form.

Step-by-step solution

Rewrite the function using trigonometric identities

Recall the following identities: \(1 - \cot x = 1 - \frac{\cos x}{\sin x}\) \(1 - \tan x = 1 - \frac{\sin x}{\cos x}\) Now, rewrite the given function in terms of trigonometric functions: \[y = \frac{\sin^2 x}{\sin x - \cos x} + \frac{\cos^2 x}{\cos x - \sin x}\] #Step 2: Calculate the derivative#

Apply the Quotient Rule

To calculate the derivative of \(y\) with respect to \(x\), apply the Quotient Rule, which states that: \(\frac{d}{dx} \left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}\) Differentiate both terms separately: For the first term: \(u_1 = \sin^2 x\) \(v_1 = \sin x - \cos x\) \(u'_1 = 2\sin x \cos x\) \(v'_1 = \cos x + \sin x\) Applying the Quotient Rule: \(\frac{d}{dx} \left(\frac{\sin^2 x}{\sin x - \cos x}\right) = \frac{(2\sin x\cos x)(\sin x - \cos x) - (\sin^2 x)(\cos x + \sin x)}{(\sin x - \cos x)^2}\) For the second term: \(u_2 = \cos^2 x\) \(v_2 = \cos x - \sin x\) \(u'_2 = -2\sin x \cos x\) \(v'_2 = -\sin x - \cos x\) Applying the Quotient Rule: \(\frac{d}{dx} \left(\frac{\cos^2 x}{\cos x - \sin x}\right) = \frac{(-2\sin x\cos x)(\cos x - \sin x) - (\cos^2 x)(-\sin x - \cos x)}{(\cos x - \sin x)^2}\) #Step 3: Combine the derivatives#

Add the derivatives of each term

Now, add both derivatives to find the overall derivative: \(\frac{dy}{dx} = \frac{(2\sin x\cos x)(\sin x - \cos x) - (\sin^2 x)(\cos x + \sin x)}{(\sin x - \cos x)^2} + \frac{(-2\sin x\cos x)(\cos x - \sin x) - (\cos^2 x)(-\sin x - \cos x)}{(\cos x - \sin x)^2}\) #Step 4: Evaluate at x = π/4#

Plug in x = π/4

Let's simplify the expression when \(x = \frac{\pi}{4}\): \(\sin\frac{\pi}{4} = \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}\) \(\frac{dy}{dx}\Big|_{x=\frac{\pi}{4}} = \frac{(2(\frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}} ) )(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) - \left(\left(\frac{1}{\sqrt{2}}\right)^2\right)(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} ) }{\left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)^2} + \frac{(-2(\frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\right))(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) - \left(\left(\frac{1}{\sqrt{2}}\right)^2\right)(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}})}{\left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)^2}\) After simplification, we get: \(\frac{dy}{dx}\Big|_{x=\frac{\pi}{4}} = \frac{0}{0} + \frac{0}{0}\) Since the expression is undefined at \(x = \frac{\pi}{4}\), we can't choose a valid answer from the given options. None of the options presented are correct due to the indeterminate form.

Key concepts

quotient rule

One of the powerful methods in calculus for differentiating rational functions is the Quotient Rule.
It's specifically designed for functions expressed as a ratio of two smaller functions, say \(f(x) = \frac{u(x)}{v(x)}\). To find the derivative of such functions, the Quotient Rule comes in handy, capturing how each part of the fraction affects the overall derivative.
The rule states:

• Differentiate the numerator \(u(x)\) to get \(u'(x)\).
• Differentiate the denominator \(v(x)\) to get \(v'(x)\).
• Then the derivative \(f'(x)\) is given by: \[f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}\]
Apply this method by keeping track of each step to ensure accuracy, especially useful in complex expressions such as those involving trigonometric identities.

trigonometric identities

Trigonometric identities are essential in simplifying complex trigonometric expressions.
They provide relationships between trigonometric functions such as sine, cosine, and tangent, enabling the transformation of expressions into more manageable forms.
A few key identities include:

• Pythagorean identity: \( \sin^2 x + \cos^2 x = 1 \)
• Reciprocal identities: \( \cot x = \frac{1}{\tan x} \)
• Co-function identities: \( \sin(90^\circ - x) = \cos x \)
In the given exercise, using identities like \(1 - \cot x = 1 - \frac{\cos x}{\sin x}\) and \(1 - \tan x = 1 - \frac{\sin x}{\cos x}\) helps simplify the function and make differentiation with the Quotient Rule more straightforward. Recognizing these identities can often transform a challenging problem into a simpler one.

indeterminate forms

Indeterminate forms occur in calculus when expressions do not initially yield a conclusive result.
Common types include \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\), which often appear when evaluating limits or when substituting values into derivatives.
These forms require special techniques to resolve, such as:

• L'Hospital's Rule: This involves differentiating the numerator and denominator separately to evaluate the limit again.
• Algebraic manipulation: Simplifying the expression by factoring or rationalizing can often provide clarity and resolve the indeterminate form.

In our exercise, substituting \(x = \frac{\pi}{4}\) into the derivative led to an \(\frac{0}{0}\) indeterminate form, indicating a need for further steps to simplify or transform the expression for practical evaluation.

JEE Main Mathematics

JEE Main Mathematics is a crucial subject for students pursuing engineering in India, testing a broad range of concepts, including calculus.
Within this field lies the test of understanding in areas like differentiating trigonometric functions, which are pivotal for solving problems similar to the exercise discussed.
The examination format often includes:

• Multiple Choice Questions (MCQs), where understanding derivative techniques like the Quotient Rule can help swiftly identify the correct answer.
• Numerical value questions, demanding precise calculations.

Concepts such as indeterminate forms and using trigonometric identities are frequently included, thus mastering these will streamline solving even the trickiest problems. Building a strong foundation in these areas through practice and concept-driven learning greatly benefits students aiming to excel in JEE Main Mathematics.