If $y = \dfrac{\sin x + \cos x}{\sin x - \cos x}$, then $\dfrac{dy}{dx}$ at $x = 0$ is (A) $-2$ (B) $0$ (C) $\dfrac{1}{2}$ (D) does not exist

> [!TLDR] > The derivative of $y = \frac{\sin x + \cos x}{\sin x - \cos x}$ at $x = 0$ is found using the quotient rule, simplifying to $\frac{dy}{dx} = -\frac{2}{(\sin x - \cos x)^2}$, which evaluates to $-2$ at $x = 0$. The key idea here is that the derivative at a specific point can be computed by first finding the general derivative function using standard rules, then plugging in the given $x$ value. For a rational function like this, the quotient rule is the natural tool — but we must be careful with signs and simplifications. Let’s work through it step by step. 1. **Identify the function and the rule needed** We have $y = \frac{u}{v}$, where $u = \sin x + \cos x$ and $v = \sin x - \cos x$. The quotient rule states: $$ \frac{dy}{dx} = \frac{v \cdot u' - u \cdot v'}{v^2} $$ 2. **Find the derivatives of $u$ and $v$** - $u' = \frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x$ - $v' = \frac{d}{dx}(\sin x - \cos x) = \cos x + \sin x$ Notice that $u'$ and $v'$ are related: $u' = -(\sin x - \cos x) = -v$, and $v' = \sin x + \cos x = u$. This symmetry will simplify things. 3. **Apply the quotient rule** $$ \frac{dy}{dx} = \frac{(\sin x - \cos x)(\cos x - \sin x) - (\sin x + \cos x)(\cos x + \sin x)}{(\sin x - \cos x)^2} $$ Look at the numerator: the first term is $(\sin x - \cos x)(\cos x - \sin x)$. Notice that $\cos x - \sin x = -(\sin x - \cos x)$, so this term becomes $-(\sin x - \cos x)^2$. The second term is $(\sin x + \cos x)(\cos x + \sin x) = (\sin x + \cos x)^2$. So the numerator simplifies to: $$ -(\sin x - \cos x)^2 - (\sin x + \cos x)^2 $$ 4. **Simplify the numerator further** Expand both squares: - $(\sin x - \cos x)^2 = \sin^2 x - 2\sin x \cos x + \cos^2 x = 1 - \sin 2x$ (since $\sin^2 x + \cos^2 x = 1$ and $2\sin x \cos x = \sin 2x$) - $(\sin x + \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x = 1 + \sin 2x$ …

If y = (sin x + cos x)/(sin x - cos x), then (dy)/ – Class 11 Mathematics

NCERT Exemplar · Q71

y = \dfrac{\sin x + \cos x}{\sin x - \cos x}

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Concept understanding — Derivative at a Point

Derivative at a Point: From Intuition to Precision

Imagine you're driving a car. Your speedometer doesn't tell you your average speed over the whole trip — it tells you your speed right now, at this exact instant. That's the core idea of a derivative at a point: it measures how fast something is changing at a single moment.

The Intuition: Instantaneous Rate of Change

Let's start with something simpler. Suppose you drop a ball from a height. The distance it has fallen after $t$ seconds is given by $s(t) = 4.9t^2$ metres (ignoring air resistance).

If I ask you "how fast was the ball falling after exactly 2 seconds?", you can't just divide distance by time — that gives an average speed over an interval. You need the speed at $t=2$ , not between $t=1$ and $t=3$ .

Here's the trick: take a very small time interval around $t=2$ , say from $t=2$ to $t=2+h$ where $h$ is tiny. The average speed over that interval is:

\frac{s(2+h) - s(2)}{h}

If $h = 0.1$ , you get one number. If $h = 0.01$ , you get a slightly different number. As $h$ gets closer and closer to 0, these average speeds approach a single value — that's the instantaneous speed at $t=2$ .

Note

This "shrinking interval" idea is the heart of the derivative. We're not setting $h=0$ (that would give $\frac{0}{0}$ , which is meaningless). We're letting $h$ approach 0 and seeing what the ratio approaches.

The Precise Definition

For a function $f(x)$ , the derivative at a point $x = a$ is defined as:

f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

provided this limit exists.

Let's break this down piece by piece:

$f(a+h) - f(a)$ is the change in the function's value when you move from $a$ to $a+h$ .
Dividing by $h$ gives the average rate of change over that interval.
Taking the limit as $h \to 0$ shrinks the interval to a single point, giving the instantaneous rate of change.

f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

Geometric Interpretation

There's also a beautiful geometric meaning. The average rate of change $\frac{f(a+h) - f(a)}{h}$ is the slope of the secant line through the points $(a, f(a))$ and $(a+h, f(a+h))$ .

As $h \to 0$ , these two points get closer together, and the secant line approaches a line that just touches the curve at $x=a$ — the tangent line. So:

The derivative at a point equals the slope of the tangent line to the curve at that point.

A Concrete Example

Let's compute the derivative of $f(x) = x^2$ at $x = 3$ .

Using the definition:

f'(3) = \lim_{h \to 0} \frac{(3+h)^2 - 3^2}{h}

Expand $(3+h)^2 = 9 + 6h + h^2$ :

f'(3) = \lim_{h \to 0} \frac{9 + 6h + h^2 - 9}{h} = \lim_{h \to 0} \frac{6h + h^2}{h}

Factor $h$ :

f'(3) = \lim_{h \to 0} \frac{h(6 + h)}{h} = \lim_{h \to 0} (6 + h)

Since $h \to 0$ , this approaches $6$ .

Important

The derivative of $x^2$ at $x=3$ is 6. This means:

At $x=3$ , the function is increasing at a rate of 6 units per unit change in $x$ .
The tangent line to $y = x^2$ at $(3, 9)$ has slope 6.

Notation

You'll see several notations for "the derivative of $f$ at $x=a$ ":

$f'(a)$ — Lagrange notation (most common) …

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The derivative of $y = \frac{\sin x + \cos x}{\sin x - \cos x}$ at $x = 0$ is found using the quotient rule, simplifying to $\frac{dy}{dx} = -\frac{2}{(\sin x - \cos x)^2}$ , which evaluates to $-2$ at $x = 0$ .

The key idea here is that the derivative at a specific point can be computed by first finding the general derivative function using standard rules, then plugging in the given $x$ value. For a rational function like this, the quotient rule is the natural tool — but we must be careful with signs and simplifications.

Let’s work through it step by step.

Identify the function and the rule needed We have $y = \frac{u}{v}$ , where $u = \sin x + \cos x$ and $v = \sin x - \cos x$ . The quotient rule states:

\frac{dy}{dx} = \frac{v \cdot u' - u \cdot v'}{v^2}

Find the derivatives of $u$ and $v$
- $u' = \frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x$
- $v' = \frac{d}{dx}(\sin x - \cos x) = \cos x + \sin x$
Notice that $u'$ and $v'$ are related: $u' = -(\sin x - \cos x) = -v$ , and $v' = \sin x + \cos x = u$ . This symmetry will simplify things.
Apply the quotient rule

\frac{dy}{dx} = \frac{(\sin x - \cos x)(\cos x - \sin x) - (\sin x + \cos x)(\cos x + \sin x)}{(\sin x - \cos x)^2}

Look at the numerator: the first term is $(\sin x - \cos x)(\cos x - \sin x)$ . Notice that $\cos x - \sin x = -(\sin x - \cos x)$ , so this term becomes $-(\sin x - \cos x)^2$ .

The second term is $(\sin x + \cos x)(\cos x + \sin x) = (\sin x + \cos x)^2$ .

So the numerator simplifies to:

-(\sin x - \cos x)^2 - (\sin x + \cos x)^2

Simplify the numerator further Expand both squares:
- $(\sin x - \cos x)^2 = \sin^2 x - 2\sin x \cos x + \cos^2 x = 1 - \sin 2x$ (since $\sin^2 x + \cos^2 x = 1$ and $2\sin x \cos x = \sin 2x$ )
- $(\sin x + \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x = 1 + \sin 2x$ …

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Q.If y=sin⁡x+cos⁡xsin⁡x−cos⁡xy = \dfrac{\sin x + \cos x}{\sin x - \cos x}y=sinx−cosxsinx+cosx​, then dydx\dfrac{dy}{dx}dxdy​ at x=0x = 0x=0 is (A) −2-2−2 (B) 000 (C) 12\dfrac{1}{2}21​ (D) does not exist

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Q.If $y = \dfrac{\sin x + \cos x}{\sin x - \cos x}$ , then $\dfrac{dy}{dx}$ at $x = 0$ is
(A) $-2$
(B) $0$
(C) $\dfrac{1}{2}$
(D) does not exist

Unlock everything free for 14 days