Q.Find the derivative of $\dfrac{\sin x + \cos x}{\sin x - \cos x}$.

Concept: Quotient Rule Differentiation

We have $y = \frac{\sin x + \cos x}{\sin x - \cos x}$.

Let $u = \sin x + \cos x$ and $v = \sin x - \cos x$.

Then $u' = \cos x - \sin x$ and $v' = \cos x + \sin x$.

Using the quotient rule $\frac{dy}{dx} = \frac{u'v - uv'}{v^2}$:

Numerator: $(\cos x - \sin x)(\sin x - \cos x) - (\sin x + \cos x)(\cos x + \sin x)$.

Notice $(\cos x - \sin x)(\sin x - \cos x) = -(\sin x - \cos x)^2$ and $(\sin x + \cos x)(\cos x + \sin x) = (\sin x + \cos x)^2$. …

The Quotient Rule is the natural choice here because we have one function divided by another. When you see a fraction of two differentiable functions, the rule says: the derivative of $\frac{u}{v}$ is $\frac{u'v - uv'}{v^2}$. The key insight is that the numerator and denominator are both simple trigonometric sums, so their derivatives are straightforward — the real work is in the algebra that follows.

Let’s work through it step by step.

1. Identify $u$ and $v$

   Let $u = \sin x + \cos x$ and $v = \sin x - \cos x$.

   Then $u' = \cos x - \sin x$ (since derivative of $\sin x$ is $\cos x$, and derivative of $\cos x$ is $-\sin x$).

   Similarly, $v' = \cos x + \sin x$ (derivative of $\sin x$ is $\cos x$, derivative of $-\cos x$ is $+\sin x$).

2. Apply the Quotient Rule

   The derivative $y'$ is:

3. Simplify the numerator — notice the pattern

   Look at the first product: $(\cos x - \sin x)(\sin x - \cos x)$.

   Factor a $-1$ from the second factor: $\sin x - \cos x = -(\cos x - \sin x)$.

   So the first product becomes $(\cos x - \sin x) \cdot [-(\cos x - \sin x)] = -(\cos x - \sin x)^2$.

   Now the second product: $(\sin x + \cos x)(\cos x + \sin x)$ is just $(\sin x + \cos x)^2$, since addition is commutative.

   So the numerator is:

4. Expand and combine Expand each square: …