If $x = e^{\cos 3t}$ and $y = e^{\sin 3t}$, prove that $\dfrac{dy}{dx} = -\dfrac{y \log x}{x \log y}$. OR Show that: $\dfrac{d}{dx}(|x|) = \dfrac{x}{|x|},\ x \ne 0$

> [!TLDR] > **Part (a):** parametric differentiation plus $\log x=\cos 3t,\ \log y=\sin 3t$ gives $\frac{dy}{dx}=-\frac{y\log x}{x\log y}$. **Part (b):** splitting into $x>0$ and $x<0$ shows $\frac{d}{dx}|x|=\frac{x}{|x|}$. ## Part (a) Both variables are functions of $t$, so $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$. 1. **Differentiate $x=e^{\cos 3t}$.** $$\frac{dx}{dt}=e^{\cos 3t}\cdot(-\sin 3t)\cdot 3=-3x\sin 3t.$$ 2. **Differentiate $y=e^{\sin 3t}$.** $$\frac{dy}{dt}=e^{\sin 3t}\cdot(\cos 3t)\cdot 3=3y\cos 3t.$$ 3. **Ratio.** $$\frac{dy}{dx}=\frac{3y\cos 3t}{-3x\sin 3t}=-\frac{y\cos 3t}{x\sin 3t}.$$ 4. **Eliminate $t$.** Taking logs of the originals, $\log x=\cos 3t$ and $\log y=\sin 3t$, so …

If x = ecos 3t and y = esin 3t, prove that (dy)/(dx) = -(y log x)/(x log y). OR Show that: (d)/(dx)(|x|) = (x)

Question

x = e^{\cos 3t}

(OR)

\dfrac{d}{dx}(|x|) = \dfrac{x}{|x|},\ x \ne 0

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Part (a)Concept understanding — Parametric Second Derivative

Parametric Second Derivative

When a curve is given parametrically as $x = x(t)$ , $y = y(t)$ , its slope is

\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{y'(t)}{x'(t)},\qquad x'(t)\neq 0.

The second derivative $\dfrac{d^2y}{dx^2}$ measures how fast that slope changes — the concavity of the path. The catch is that $\frac{dy}{dx}$ comes out as a function of $t$ , but we need its rate of change with respect to $x$ .

The key idea

Differentiate the slope with respect to $t$ , then convert that $t$ -derivative into an $x$ -derivative by dividing by $dx/dt$ (chain rule):

\frac{d^2y}{dx^2}=\frac{d}{dx}\!\left(\frac{dy}{dx}\right)=\frac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}.

Carrying this out with the quotient rule gives a compact closed form:

\frac{d^2y}{dx^2}=\frac{x'(t)\,y''(t)-y'(t)\,x''(t)}{\big[x'(t)\big]^3}.

Watch out

Do not write $\dfrac{d^2y}{dx^2}=\dfrac{d^2y/dt^2}{d^2x/dt^2}$ . The parametric second derivative is not the ratio of the second $t$ -derivatives — that tempting shortcut is wrong.

Worked illustration

For the cycloid $x=t-\sin t$ , $y=1-\cos t$ :

First derivatives: $\dfrac{dx}{dt}=1-\cos t$ , $\dfrac{dy}{dt}=\sin t$ , so $\dfrac{dy}{dx}=\dfrac{\sin t}{1-\cos t}$ .
Differentiate $\dfrac{dy}{dx}$ with respect to $t$ , then divide by $\dfrac{dx}{dt}=1-\cos t$ , which simplifies to $\frac{d^2y}{dx^2}=-\frac{1}{(1-\cos t)^2}.$ …

Part (b)Concept understanding — Differentiability of Absolute Value

Differentiability of the Absolute Value Function

Start with something familiar: the absolute value of $x$ , written $|x|$ , is its distance from zero on the number line. So $|3| = 3$ , $|-5| = 5$ , and $|0| = 0$ . Graphically, it looks like a V-shape — two straight lines meeting at the origin.

Differentiability is about whether a function has a well-defined slope (derivative) at a point. For smooth curves like $x^2$ or $\sin x$ , the slope exists everywhere. But the absolute value function has a sharp corner at $x = 0$ — and that corner is the whole story.

Intuition: Why the corner matters

Walk along $y = |x|$ from left to right. Approaching $x = 0$ from the left, the slope is $-1$ (the line goes downward). Leaving $x = 0$ to the right, the slope is suddenly $+1$ (the line goes upward). At $x = 0$ , there's no single slope — it changes abruptly. That's why $|x|$ is not differentiable at $x = 0$ . Everywhere else — for $x < 0$ and $x > 0$ — the graph is a straight line with constant slope, so $|x|$ is differentiable at every point except $x = 0$ .

Note

A function must be continuous to be differentiable, but continuity alone isn't enough. The absolute value function is continuous at $x = 0$ (no break), yet fails to be differentiable there because of the sharp corner.

The precise statement

Let $f(x) = |x|$ . Then:

For $x > 0$ : $f(x) = x$ , so $f'(x) = 1$ .
For $x < 0$ : $f(x) = -x$ , so $f'(x) = -1$ .
At $x = 0$ : the derivative does not exist, because the left-hand and right-hand derivatives are different numbers.

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

Watch out

A common mistake is to think that because $|x|$ is continuous at $x=0$ , it must be differentiable there. Continuity is necessary for differentiability, but not sufficient. The absolute value function is the classic counterexample.

The bigger picture …

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Part (a): parametric differentiation plus $\log x=\cos 3t,\ \log y=\sin 3t$ gives $\frac{dy}{dx}=-\frac{y\log x}{x\log y}$ . Part (b): splitting into $x>0$ and $x<0$ shows $\frac{d}{dx}|x|=\frac{x}{|x|}$ .

Part (a)

Both variables are functions of $t$ , so $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ .

Differentiate $x=e^{\cos 3t}$ .

\frac{dx}{dt}=e^{\cos 3t}\cdot(-\sin 3t)\cdot 3=-3x\sin 3t.

Differentiate $y=e^{\sin 3t}$ .

\frac{dy}{dt}=e^{\sin 3t}\cdot(\cos 3t)\cdot 3=3y\cos 3t.

Ratio.

\frac{dy}{dx}=\frac{3y\cos 3t}{-3x\sin 3t}=-\frac{y\cos 3t}{x\sin 3t}.

Eliminate $t$ . Taking logs of the originals, $\log x=\cos 3t$ and $\log y=\sin 3t$ , so …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 17 on this concept.

CBSE 2026Set V11 mark
$[0, 3, -1, 2, -2, 1]$
›Reveal solutionSolution

Just left of $0$ , $|x|=-x$ has slope $-1$ , so the left-hand derivative is $-1$ .

The left-hand derivative at $x=0$ is

$\lim_{h\to0^-}\frac{|0+h|-|0|}{h}=\lim_{h\to0^-}\frac{-h}{h}=-1,$ …
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CBSE 2026Set ANNUAL1 markMCQ
$x = a\cos\theta$
(a) $-\cot\theta$
(b) $\tan\theta$
(c) $a\tan\theta$
(d) None of these
›Reveal solutionSolution

For a parametric curve, $\dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx/d\theta}$ .

$x = a\cos\theta \Rightarrow \dfrac{dx}{d\theta} = -a\sin\theta$

$y = a\sin\theta \Rightarrow \dfrac{dy}{d\theta} = a\cos\theta$

…

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CBSE 2026Set ANNUAL1 mark
$Q.If x=a sec \theta and y=b tan \theta, then find \frac{dy}{dx}.$
›Reveal solutionSolution
$\dfrac{dy}{dx}=\dfrac{b}{a}\csc\theta$ .

Concept. For a curve given parametrically as $x=f(\theta),\ y=g(\theta)$ , $\dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}$ .

Steps.

$x=a\sec\theta\Rightarrow \dfrac{dx}{d\theta}=a\sec\theta\tan\theta$ .

$y=b\tan\theta\Rightarrow \dfrac{dy}{d\theta}=b\sec^2\theta$ .

$\dfrac{dy}{dx}=\dfrac{b\sec^2\theta}{a\sec\theta\tan\theta}=\dfrac{b\sec\theta}{a\tan\theta}$ . …
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CBSE 2025Set 65/1/11 markMCQ
$f(x) = |x| + |x-1|$
›Reveal solutionSolution
The function $f(x) = |x| + |x-1|$ is a sum of two absolute value functions, each continuous everywhere but with a corner at its respective critical point. At $x=0$ and $x=1$ , the left and right derivatives differ, so $f$ is continuous but not differentiable at both points — option (C).

The key to this problem is understanding what absolute value does to differentiability. The function $|x|$ has a V-shaped graph — it is continuous everywhere, but at $x=0$ the slope changes abruptly from $-1$ to $+1$ . That sharp corner means the derivative does not exist at $x=0$ , even though the function is perfectly continuous there. The same logic applies to $|x-1|$ at $x=1$ .

When you add two such functions, the sum inherits the continuity of each piece. But at a point where either term has a corner, the sum may also have a corner — unless the slopes happen to cancel, which they do not here.

Let’s check each point carefully.

Continuity at $x=0$ Compute the left-hand limit, right-hand limit, and the function value. For $x < 0$ : $|x| = -x$ , $|x-1| = -(x-1) = 1-x$ , so

$f(x) = -x + (1-x) = 1 - 2x.$
As $x \to 0^-$ , $f(x) \to 1$ .

For $x > 0$ but $x < 1$ : $|x| = x$ , $|x-1| = 1-x$ , so
$f(x) = x + (1-x) = 1.$
As $x \to 0^+$ , $f(x) \to 1$ .

Also $f(0) = |0| + |0-1| = 0 + 1 = 1$ .

Since left limit = right limit = function value, $f$ is continuous at $x=0$ .

Differentiability at $x=0$ The left-hand derivative uses the expression for $x<0$ : $f(x) = 1 - 2x$ , so $f'(x) = -2$ . Hence

$f'_-(0) = -2.$
The right-hand derivative uses the expression for $0 < x < 1$ : $f(x) = 1$ , so $f'(x) = 0$ . Hence
$f'_+(0) = 0.$
Since $-2 \neq 0$ , the left and right derivatives are different. Therefore $f$ is not differentiable at $x=0$ .

Watch out

A common mistake is to think that because $|x|$ alone is not differentiable at $0$ , the sum must also fail — which is true here, but you must check the actual slopes. If the slopes from both terms happened to match on both sides, the sum could become differentiable. Always compute the left and right derivatives explicitly. …
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CBSE 2025Set X11 markMCQ
$y = f(x)$
(a) Statement 1 is true, Statement 2 is false
(b) Statement 1 is false, Statement 2 is true
(c) Both Statements 1 and 2 are true
(d) Both Statements 1 and 2 are false
›Reveal solutionSolution

One-sided derivatives / differentiability at a corner — correct option (a).

The left-hand derivative is the slope of the left branch from $(0,1)$ to $(1,0)$ : $\dfrac{0-1}{1-0} = -1$ , so Statement 1 is true. The right branch rises from $(1,0)$ to $(2,1)$ with slope $\dfrac{1-0}{2-1} = +1$ . Since the left and right derivatives differ ( $-1 \neq +1$ ), the …
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CBSE 2025Set ANNUAL1 markMCQ
$f(x) = |x|$
(a) continuous but not differentiable
(b) differentiable but not continuous
(c) continuous and differentiable
(d) neither continuous nor differentiable
›Reveal solutionSolution

|x| has a 'corner' at x = 0 — no jump (continuous) but a sharp change in slope (not differentiable).

Continuity: $\lim_{x\to0^-}|x| = 0$ , $\lim_{x\to0^+}|x|=0$ , and $f(0)=0$ . All three agree, so f is continuous at x = 0.

Differentiability: Left-hand derivative: $\lim_{h\to0^-}\dfrac{|0+h|-|0|}{h} = \lim_{h\to0^-}\dfrac{-h}{h} = -1$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$x = a \sec\theta$
(a) $\dfrac{b}{a} \sec\theta$
(b) $\dfrac{b}{a} \text{cosec}\,\theta$
(c) $\dfrac{b}{a} \cot\theta$
(d) none of these
›Reveal solutionSolution

Differentiate x and y separately with respect to the parameter θ, then divide.

$x = a\sec\theta \Rightarrow \dfrac{dx}{d\theta} = a\sec\theta\tan\theta$

$y = b\tan\theta \Rightarrow \dfrac{dy}{d\theta} = b\sec^2\theta$

…

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CBSE 2024Set ANNUAL1 mark
$\dfrac{dy}{dx}$
›Reveal solutionSolution

Use parametric differentiation: $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ .

Given $x = 2at^2$ , $y = at^4$ .

$\frac{dx}{dt} = 4at, \qquad \frac{dy}{dt} = 4at^3$

So: …

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CBSE 2023Set 65/2/11 markMCQ
$f(x) = x|x|$
(a) continuous and differentiable at $x = 0$
(b) continuous but not differentiable at $x = 0$
(c) differentiable but not continuous at $x = 0$
(d) neither differentiable nor continuous at $x = 0$
›Reveal solutionSolution
The absolute value creates a piecewise definition, but the square in $f(x) = x|x|$ smooths out the corner that usually appears at the origin; both continuity and differentiability hold at $x = 0$ .

Understanding Differentiability of Absolute Value Functions

The absolute value function $|x|$ itself has a sharp corner at $x = 0$ , making it continuous but not differentiable there. However, when we multiply $x$ by $|x|$ , we're creating something different. The key insight is to rewrite $f(x)$ in piecewise form and check whether the pieces "join smoothly" at the origin.

Start by recalling that $|x| = x$ when $x \geq 0$ and $|x| = -x$ when $x < 0$ . This gives us:
$f(x) = x|x| = \begin{cases} x \cdot x = x^2 & \text{if } x \geq 0 \\ x \cdot (-x) = -x^2 & \text{if } x < 0 \end{cases}$
Notice that both pieces are parabolas, one opening upward and one downward, meeting at the origin.

Checking Continuity at $x = 0$

Compute the left-hand limit:

$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x^2) = 0$

Compute the right-hand limit:

$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x^2 = 0$

Evaluate at the point:

$f(0) = 0 \cdot |0| = 0$
Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 0$ , the function is continuous at $x = 0$ .

Checking Differentiability at $x = 0$

Differentiability requires that the derivative from the left equals the derivative from the right. We use the definition of the derivative:

Left-hand derivative: $f'_-(0) = \lim_{h \to 0^-} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-h^2 - 0}{h} = \lim_{h \to 0^-} \frac{-h^2}{h} = \lim_{h \to 0^-} (-h) = 0$ …
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CBSE 2023Set ANNUAL1 markMCQ
$Q. The function f(x) = |x| for all x \in R is -$
(a) continuous and differentiable at x = 0
(b) continuous but not differentiable at x = 0
(c) differentiable but not continuous at x = 0
(d) neither continuous nor differentiable at x = 0
›Reveal solutionSolution

The modulus function has no break in its graph (continuous everywhere) but has a sharp corner at $x=0$ where the left-hand and right-hand derivatives differ.

Continuity at $x=0$ : $\lim_{x\to 0^-}|x| = 0$ , $\lim_{x\to0^+}|x|=0$ , and $f(0)=0$ . All three agree, so $f$ is continuous at $x=0$ (and everywhere else, being a composition of continuous functions).

Differentiability at $x=0$ : …

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CBSE 2022Set ANNUAL1 markMCQ
$x = a\cos\theta$
(a) $\tan\theta$
(b) $-\cot\theta$
(c) $-\tan\theta$
(d) $\sec^2\theta$
›Reveal solutionSolution

Parametric differentiation gives $\frac{a\cos\theta}{-a\sin\theta}=-\cot\theta$ .

$\frac{dx}{d\theta}=-a\sin\theta$ , $\frac{dy}{d\theta}=a\cos\theta$ .

…

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CBSE 2022Set ANNUAL1 markMCQ
$x = at^2$
(a) $t$
(b) $\frac{1}{t}$
(c) $t^2$
(d) None of these
›Reveal solutionSolution

Parametric: $\frac{dy/dt}{dx/dt}=\frac{2a}{2at}=\frac1t$ .

$\frac{dx}{dt}=2at$ , $\frac{dy}{dt}=2a$ .

…

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Q.If x=ecos⁡3tx = e^{\cos 3t}x=ecos3t and y=esin⁡3ty = e^{\sin 3t}y=esin3t, prove that dydx=−ylog⁡xxlog⁡y\dfrac{dy}{dx} = -\dfrac{y \log x}{x \log y}dxdy​=−xlogyylogx​.

Part (a)

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Q.If $x = e^{\cos 3t}$ and $y = e^{\sin 3t}$ , prove that $\dfrac{dy}{dx} = -\dfrac{y \log x}{x \log y}$ .

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