Q.59 If f(x) = (sin x + cos x) / (sin x − cos x), the value of f′(x) at x = 0 is ____________.

Introduction

Finding the derivative of the function

f(x) = (sin x + cos x)/(sin x – cos x)

at x = 0 is a standard problem in calculus and often appears in IIT JAM and other competitive exams. The challenge lies in applying the quotient rule correctly and simplifying trigonometric expressions.

Step-by-Step Derivative Calculation

1. Define Numerator and Denominator

Let:

• u = sin x + cos x → u’ = cos x – sin x
• v = sin x – cos x → v’ = cos x + sin x

2. Apply the Quotient Rule

f'(x)=u’v – uv’/v²

Substituting:

f'(x)=(cosx – sinx)(sinx – cosx) – (sinx + cosx)(cosx + sinx)/(sinx – cosx)²

Simplifying the Expression

Expand both products in the numerator:

(cosx – sinx)(sinx – cosx) = sinxcosx – cos²x – sin²x + sinxcosx

(sinx + cosx)(cosx + sinx) = sinxcosx + sin²x + cos²x + sinxcosx

Combine like terms using the trigonometric identity:

sin²x + cos²x = 1

Final simplified numerator:

-2(sin²x + cos²x) = -2

Thus,

f'(x) = -2/(sinx – cosx)²

Evaluate at x = 0

Compute denominator:

(sin0 – cos0)² = (0 – 1)² = 1

Therefore:

f'(0) = -2

Final Answer

f'(0) = -2

Exam Tips

• Watch for points where the function is undefined: sin x = cos x → x = π/4 + kπ
• A removable discontinuity can still yield a valid derivative
• Recognize identities early to simplify messy algebra
• Practice similar derivatives like tan x / (1 + sec x)

Conclusion

The derivative of (sin x + cos x)/(sin x – cos x) at x = 0 evaluates to -2.

This is a great example of how applying the quotient rule and simplifying trigonometric identities can lead to a clean result.