NCERT Class 12 Mathematics — Continuity and Differentiability, Exercise 5.6. All 11 questions solved.

When $x$ and $y$ are both given in terms of a third variable $t$, the chain rule gives

$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \qquad \frac{dx}{dt} \ne 0$$

Differentiate each of $x$ and $y$ with respect to the parameter, then divide. There is no need to eliminate $t$ — which is often impossible anyway.

Key insight. Almost every answer in this exercise simplifies dramatically, and the simplification is where the marks are. Question 6 gives $\dfrac{-a\sin\theta}{a(1-\cos\theta)}$, which looks like a dead end until you use the half-angle forms $\sin\theta = 2\sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2}$ and $1 – \cos\theta = 2\sin^2\tfrac{\theta}{2}$ — and it collapses to $-\cot\tfrac{\theta}{2}$.

Question 10 is the other pattern worth knowing: both derivatives contain a common factor of $a\theta$, and cancelling it leaves $\tan\theta$ in one step.

If $x$ and $y$ are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find $\dfrac{dy}{dx}$.

Question 1

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

Solution.

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

$$\frac{dy}{dx} = \frac{4at^3}{4at} = t^2$$

Question 2

$x = a\cos\theta$, $y = b\cos\theta$

Solution.

$$\frac{dx}{d\theta} = -a\sin\theta, \qquad \frac{dy}{d\theta} = -b\sin\theta$$

The $\sin\theta$ cancels, leaving a constant — which makes sense, since $y = \tfrac{b}{a}x$ is a straight line through the origin.

Question 3

$x = \sin t$, $y = \cos 2t$

Solution.

$$\frac{dx}{dt} = \cos t, \qquad \frac{dy}{dt} = -2\sin 2t = -4\sin t\cos t$$

$$\frac{dy}{dx} = \frac{-4\sin t\cos t}{\cos t}$$

Expanding $\sin 2t$ before dividing is what makes the cancellation visible.

Question 4

$x = 4t$, $y = \dfrac{4}{t}$

Solution.

$$\frac{dx}{dt} = 4, \qquad \frac{dy}{dt} = -\frac{4}{t^2}$$

Question 5

$x = \cos\theta – \cos 2\theta$, $y = \sin\theta – \sin 2\theta$

Solution.

$$\frac{dx}{d\theta} = -\sin\theta + 2\sin 2\theta, \qquad \frac{dy}{d\theta} = \cos\theta – 2\cos 2\theta$$

Question 6

$x = a(\theta – \sin\theta)$, $y = a(1 + \cos\theta)$

Solution.

$$\frac{dx}{d\theta} = a(1 – \cos\theta), \qquad \frac{dy}{d\theta} = -a\sin\theta$$

$$\frac{dy}{dx} = \frac{-a\sin\theta}{a(1-\cos\theta)} = \frac{-2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}}$$

The half-angle identities turn this into a single cotangent.

Question 7

$x = \dfrac{\sin^3t}{\sqrt{\cos 2t}}$, $y = \dfrac{\cos^3t}{\sqrt{\cos 2t}}$

Solution. Both are quotients with the same denominator, so differentiate each with the quotient rule and note that the $\left(\cos 2t\right)$ terms in the two results are proportional.

$$\frac{dx}{dt} = \frac{3\sin^2t\cos t\sqrt{\cos 2t} + \frac{\sin^3t\sin 2t}{\sqrt{\cos 2t}}}{\cos 2t}$$

$$\frac{dy}{dt} = \frac{-3\cos^2t\sin t\sqrt{\cos 2t} + \frac{\cos^3t\sin 2t}{\sqrt{\cos 2t}}}{\cos 2t}$$

Multiplying both through by $\dfrac{\cos 2t\sqrt{\cos 2t}}{\sin^2t\cos^2t}$ and using $\cos 2t = \cos^2t – \sin^2t$ and $\sin 2t = 2\sin t\cos t$, the ratio reduces to $-\cot 3t$.

Question 8

$x = a\left(\cos t + \log\tan\dfrac{t}{2}\right)$, $y = a\sin t$

Solution. The logarithm differentiates to

$$\frac{d}{dt}\log\tan\frac{t}{2} = \frac{\sec^2\frac{t}{2}}{2\tan\frac{t}{2}} = \frac{1}{2\sin\frac{t}{2}\cos\frac{t}{2}} = \frac{1}{\sin t}$$

So

$$\frac{dx}{dt} = a\left(-\sin t + \frac{1}{\sin t}\right) = a \cdot \frac{1 – \sin^2t}{\sin t} = \frac{a\cos^2t}{\sin t}$$

and $\dfrac{dy}{dt} = a\cos t$. Dividing:

$$\frac{dy}{dx} = \frac{a\cos t\sin t}{a\cos^2t}$$

Question 9

$x = a\sec\theta$, $y = b\tan\theta$

Solution.

$$\frac{dx}{d\theta} = a\sec\theta\tan\theta, \qquad \frac{dy}{d\theta} = b\sec^2\theta$$

$$\frac{dy}{dx} = \frac{b\sec^2\theta}{a\sec\theta\tan\theta} = \frac{b}{a} \cdot \frac{\sec\theta}{\tan\theta} = \frac{b}{a} \cdot \frac{1}{\sin\theta}$$

Question 10

$x = a(\cos\theta + \theta\sin\theta)$, $y = a(\sin\theta – \theta\cos\theta)$

Solution. Each derivative needs the product rule on the second term, and in both cases two terms cancel:

$$\frac{dx}{d\theta} = a\left(-\sin\theta + \sin\theta + \theta\cos\theta\right) = a\theta\cos\theta$$

$$\frac{dy}{d\theta} = a\left(\cos\theta – \cos\theta + \theta\sin\theta\right) = a\theta\sin\theta$$

The common factor $a\theta$ cancels on division.

Question 11

If $x = \sqrt{a^{\sin^{-1}t}}$, $y = \sqrt{a^{\cos^{-1}t}}$, show that $\dfrac{dy}{dx} = -\dfrac{y}{x}$.

Solution. Write $x = a^{\frac12\sin^{-1}t}$ and $y = a^{\frac12\cos^{-1}t}$, and take logarithms:

$$\log x = \frac12\sin^{-1}t\log a, \qquad \log y = \frac12\cos^{-1}t\log a$$

Differentiating with respect to $t$:

$$\frac{1}{x}\frac{dx}{dt} = \frac{\log a}{2\sqrt{1-t^2}}, \qquad \frac{1}{y}\frac{dy}{dt} = \frac{-\log a}{2\sqrt{1-t^2}}$$

So $\dfrac{dx}{dt} = \dfrac{x\log a}{2\sqrt{1-t^2}}$ and $\dfrac{dy}{dt} = \dfrac{-y\log a}{2\sqrt{1-t^2}}$. Dividing, everything except the ratio cancels:

$$\frac{dy}{dx} = -\frac{y}{x}$$

The result also follows from $\sin^{-1}t + \cos^{-1}t = \tfrac{\pi}{2}$, which makes $xy$ constant — and a constant product always gives $\tfrac{dy}{dx} = -\tfrac{y}{x}$.

Common mistakes

• Dividing the wrong way round. It is $\tfrac{dy/dt}{dx/dt}$, not the reciprocal. The units check: $\tfrac{dy}{dx}$ must have $y$ on top.
• Stopping before simplifying. Questions 6, 7, 8 and 9 all have answers that look like nothing until a trigonometric identity is applied.
• Forgetting the product rule. Question 10’s $\theta\sin\theta$ and $\theta\cos\theta$ each need it, and the cancellation only happens if it is done correctly.
• Misdifferentiating $\log\tan\tfrac{t}{2}$. The chain rule gives $\tfrac{\sec^2(t/2)}{2\tan(t/2)}$, which simplifies to $\tfrac{1}{\sin t}$ — a result worth remembering, as it recurs.
• Trying to eliminate the parameter. The question explicitly says not to, and in questions 7 and 8 it is not practical anyway.
• Dividing by zero. The formula requires $\tfrac{dx}{dt} \ne 0$; in question 3 that excludes $t = \tfrac{\pi}{2}$.

Practise next

• Exercise 5.7 — second order derivatives, including the parametric case.
• Exercise 5.5 — the logarithmic differentiation used in question 11.