Q.Find $\frac{dy}{dx}$ in the following: $x = \cos \theta - \cos 2\theta, y = \sin \theta - \sin 2\theta$

Showing the 12 most recent of 16 on this concept.

1. KCET 2022Set C-41 markMCQ
   Q.If $x = e^{\theta} \sin \theta$, $y = e^{\theta} \cos \theta$ where $\theta$ is a parameter, then $\frac{dy}{dx}$ at (1, 1) is equal to (A) $\frac{1}{2}$ (B) $-\frac{1}{2}$ (C) $-\frac{1}{4}$ (D) $0$

   ›Reveal solutionSolution

   Use $\dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}$; at the point $(1,1)$ we have $\sin\theta=\cos\theta$, which kills the numerator, so the derivative is $0$.

   Step 1 — Why parametric differentiation

   Both $x$ and $y$ are given in terms of a third variable $\theta$, not of each other. The chain rule then gives

   $\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}\qquad\left(\frac{dx}{d\theta}\neq 0\right)$

   Step 2 — Differentiate each with the product rule

   $x=e^{\theta}\sin\theta \;\Rightarrow\; \frac{dx}{d\theta}=e^{\theta}\sin\theta+e^{\theta}\cos\theta=e^{\theta}(\sin\theta+\cos\theta)$

   $y=e^{\theta}\cos\theta \;\Rightarrow\; \frac{dy}{d\theta}=e^{\theta}\cos\theta-e^{\theta}\sin\theta=e^{\theta}(\cos\theta-\sin\theta)$

   Step 3 — Form the ratio

   $\frac{dy}{dx}=\frac{e^{\theta}(\cos\theta-\sin\theta)}{e^{\theta}(\sin\theta+\cos\theta)}=\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}$

   The factor $e^{\theta}$ (never zero) cancels — this is the whole point of taking $e^{\theta}$ common.

   Step 4 — Impose the condition of the point $(1,1)$

   At that point $x=y$, so …

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2. COMEDK 2024Set 2024-A1 markMCQ
   Q.

   [!FORMULA] $\text { If } x^2+y^2=t+\frac{1}{t} \text { and } x^4+y^4=t^2+\frac{1}{t^2} \text { then } \frac{d y}{d x}=$

   (A) $\frac{x}{2 y}$ (B) $-\frac{y}{x}$ (C) $-\frac{x}{2 y}$ (D) $\frac{y}{x}$

   ›Reveal solutionSolution

   The key is to notice that the given equations imply a simple relation between $x$ and $y$: $x^2 + y^2 = t + 1/t$ and $x^4 + y^4 = t^2 + 1/t^2$ force $x^2 y^2 = 1$. Differentiating implicitly gives $dy/dx = -y/x$, so the answer is (B).

   We start with two parametric-looking equations in $x, y, t$:

   $x^2 + y^2 = t + \frac{1}{t}, \qquad x^4 + y^4 = t^2 + \frac{1}{t^2}.$

   The goal is to find $\frac{dy}{dx}$ without needing $t$ explicitly — we want a direct relation between $x$ and $y$.

   ---

   1. Spot the algebraic structure

   Notice that $t^2 + 1/t^2$ is the square of $t + 1/t$ minus 2:

   $\left(t + \frac{1}{t}\right)^2 = t^2 + 2 + \frac{1}{t^2} \quad\Rightarrow\quad t^2 + \frac{1}{t^2} = \left(t + \frac{1}{t}\right)^2 - 2.$

   So the second equation becomes:

   $x^4 + y^4 = (x^2 + y^2)^2 - 2.$

   ---

   2. Expand and simplify

   Expand $(x^2 + y^2)^2$:

   $(x^2 + y^2)^2 = x^4 + 2x^2 y^2 + y^4.$

   Thus:

   $x^4 + y^4 = x^4 + 2x^2 y^2 + y^4 - 2.$

   Cancel $x^4 + y^4$ from both sides, leaving:

   $0 = 2x^2 y^2 - 2 \quad\Rightarrow\quad x^2 y^2 = 1.$

   Tip

   This is the hidden gem: the parameter $t$ cancels completely, leaving a simple hyperbola-like relation $x^2 y^2 = 1$, i.e. $xy = \pm 1$.

   ---

   3. Differentiate implicitly

   From $x^2 y^2 = 1$, differentiate both sides with respect to $x$:

   $\frac{d}{dx}(x^2 y^2) = \frac{d}{dx}(1) = 0.$

   Use the product rule (or treat as $(x^2)(y^2)$): …

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3. COMEDK 2021Set 2021-B1 markMCQ
   Q.If $y = \sqrt{\sin x + y}$, then dy/dx = (A) $\frac{\cos x}{2y - 1}$ (B) $\frac{\cos x}{1 - 2y}$ (C) $\frac{2y - 1}{\cos x}$ (D) $\frac{1 - 2y}{\cos x}$

   ›Reveal solutionSolution

   $\dfrac{dy}{dx}=\dfrac{\cos x}{2y-1}$.

   Given $y=\sqrt{\sin x+y}$, square both sides: $y^2=\sin x+y$.

   Differentiate implicitly w.r.t. $x$:

   $2y\frac{dy}{dx}=\cos x+\frac{dy}{dx}\implies (2y-1)\frac{dy}{dx}=\cos x.$ …

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4. COMEDK 2024Set 2024-E1 markMCQ
   Q.

   [!FORMULA] $\text { If } y=\sqrt{\sin x+y} \text { then find } \frac{d y}{d x} \text { at } x=0, \quad y=1$

   (A) 0 (B) 1 (C) 2 (D) $-$1

   ›Reveal solutionSolution

   The equation defines $y$ implicitly; we differentiate both sides, substitute $x=0, y=1$, and solve for $\frac{dy}{dx}$ to get $0$.

   We are given $y = \sqrt{\sin x + y}$. This is not an explicit function $y(x)$ in the usual sense because $y$ appears on both sides. The key is to treat it as an implicit relation between $x$ and $y$. We differentiate both sides with respect to $x$, remembering that $y$ is a function of $x$, and then plug in the given point $(x=0, y=1)$ to find the slope.

   1. Rewrite the equation to avoid the square root for easier differentiation. Square both sides:

   $y^2 = \sin x + y$

   This is valid because $y = \sqrt{\cdots}$ implies $y \ge 0$, and at $(0,1)$ it's fine.

   2. Differentiate implicitly with respect to $x$:

   $\frac{d}{dx}(y^2) = \frac{d}{dx}(\sin x) + \frac{d}{dx}(y)$

   Using the chain rule on $y^2$ gives $2y \frac{dy}{dx}$, and $\frac{d}{dx}(\sin x) = \cos x$, and $\frac{d}{dx}(y) = \frac{dy}{dx}$.

   So:

   $2y \frac{dy}{dx} = \cos x + \frac{dy}{dx}$

   3. Solve for $\frac{dy}{dx}$ algebraically. Bring terms involving $\frac{dy}{dx}$ to one side:

   $2y \frac{dy}{dx} - \frac{dy}{dx} = \cos x$

   Factor out $\frac{dy}{dx}$:

   $\frac{dy}{dx} (2y - 1) = \cos x$

   Hence:

   $\frac{dy}{dx} = \frac{\cos x}{2y - 1}$

   4. Substitute the given values $x=0$, $y=1$: …

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5. COMEDK 2022Set 20221 markMCQ
   Q.If the tangent to the curve $xy + ax + by = 0$ at (1, 1) is inclined at an angle ${\tan ^{ - 1}}2$ with X-axis, then (A) $a = 1,b = 2$ (B) $a = 1,b = - 2$ (C) $a = - 1,b = 2$ (D) $a = - 1,b = - 2$

   ›Reveal solutionSolution

   Check: curve xy + x - 2y = 0 through (1,1): 1 + 1 - 2 = 0. Slope = -(1+1)/(1-2) = -2/-1 = 2. Correct.

   Concept: Implicit differentiation + slope of tangent = tan(theta).

   Curve: xy + ax + by = 0 passes through (1,1):

   1*1 + a(1) + b(1) = 0 => a + b = -1 ... (i)

   Differentiate implicitly:

   y + x y' + a + b y' = 0

   y'(x + b) = -(y + a)

   y' = -(y + a)/(x + b)

   At (1,1): y' = -(1 + a)/(1 + b)

   The tangent is inclined at angle arctan(2), so slope = 2:

   -(1 + a)/(1 + b) = 2

   From (i), b = -1 - a, so 1 + b = -a. Substituting: …

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6. COMEDK 2024Set 2024-M1 markMCQ
   Q.

   [!FORMULA] $\text { If } \sin y=x(\cos (a+y)) \text {, then find } \frac{d y}{d x} \text { when } x=0$

   (A) 1 (B) sec a (C) cos a (D) $-1$

   ›Reveal solutionSolution

   Differentiate implicitly (easiest via $x=\sin y/\cos(a+y)$); at $x=0,\;y=0$ the derivative reduces to $\cos a$ (option C).

   Given $\sin y=x\cos(a+y)$. When $x=0$, $\sin y=0\Rightarrow y=0$.

   Solve for $x$ and differentiate with respect to $y$:

   $x=\frac{\sin y}{\cos(a+y)}$

   $\frac{dx}{dy}=\frac{\cos y\cos(a+y)-\sin y\big(-\sin(a+y)\big)}{\cos^2(a+y)}=\frac{\cos y\cos(a+y)+\sin y\sin(a+y)}{\cos^2(a+y)}$

   The numerator is $\cos\big((a+y)-y\big)=\cos a$, so …

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7. KCET 2020Set A-11 markMCQ
   Q.If $(xe)^y = e^x$, then $\frac{dy}{dx}$ is (A) $\frac{\log x}{(1 + \log x)^2}$ (B) $\frac{1}{(1 + \log x)^2}$ (C) $\frac{\log x}{(1 + \log x)}$ (D) $\frac{e^x}{x(y-1)}$

   ›Reveal solutionSolution

   Logarithmic differentiation: take $\log$ of both sides to free $y$ from the exponent, solve for $y$ explicitly, then differentiate.

   Step 1 — Take natural logarithms (why: $y$ sits in an exponent, and $\log$ brings it down).

   $(xe)^y = e^x \;\Longrightarrow\; y\,\log(xe) = x\log e = x.$

   Step 2 — Simplify $\log(xe)$.

   $\log(xe) = \log x + \log e = \log x + 1.$

   So

   $y(1+\log x) = x \qquad\Longrightarrow\qquad y = \frac{x}{1+\log x}.$

   Step 3 — Differentiate with the quotient rule.

   With $u = x,\; v = 1+\log x$, we have $u' = 1$ and $v' = \dfrac{1}{x}$:

   $\frac{dy}{dx} = \frac{v\,u' - u\,v'}{v^{2}} = \frac{(1+\log x)(1) - x\cdot\frac{1}{x}}{(1+\log x)^{2}}.$

   Step 4 — Simplify. …

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8. COMEDK 2021Set 20211 markMCQ
   Q.The equation of normal to the curve $y = {(1 + x)^y} + {\sin ^{ - 1}}({\sin ^2}x)$ at $x = 0$ is (A) $x + y = 1$ (B) $x - y = 1$ (C) $x + y = - 1$ (D) $x - y = - 1$

   ›Reveal solutionSolution

   Step 3 - the normal. Slope of tangent m = 1 -> slope of normal = -1/m = -1. Normal through (0, 1): y - 1 = -1 (x - 0) y - 1 = -x x + y = 1

   Concept: Equation of the normal - find the point, find dy/dx (implicit differentiation), then normal slope = -1/(dy/dx).

   Curve: y = (1 + x)^y + sin^(-1)(sin^2 x)

   Step 1 - the point at x = 0:

   y = (1 + 0)^y + sin^(-1)(sin^2 0) = 1 + sin^(-1)(0) = 1 + 0 = 1

   So the point is (0, 1).

   Step 2 - differentiate.

   Let u = (1 + x)^y. Then log u = y log(1 + x), and

   (1/u) du/dx = y' log(1 + x) + y/(1 + x)

   du/dx = (1 + x)^y [ y' log(1 + x) + y/(1 + x) ]

   At x = 0, y = 1: (1+0)^1 = 1, log(1) = 0, so du/dx | 0 = 1 * [ y'(0) + 1/1 ] = 1.

   (The y' log(1+x) term vanishes because log 1 = 0.)

   For the second term, v = sin^(-1)(sin^2 x): …

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9. COMEDK 2023Set 2023-M1 markMCQ
   Q.The slope of the tangent to the curve, $y=x^2-x y$ at $\left(1, \frac{1}{2}\right)$ is (A) $\frac{4}{3}$ (B) $\frac{2}{3}$ (C) $\frac{3}{4}$ (D) $\frac{3}{2}$

   ›Reveal solutionSolution

   Implicit differentiation of $y=x^2-xy$ gives a slope of $3/4$ at the point $(1,\tfrac12)$.

   Differentiate $y = x^2 - xy$ with respect to $x$ (product rule on $xy$):

   $\frac{dy}{dx} = 2x - \left(y + x\frac{dy}{dx}\right).$

   Collect the derivative terms:

   $\frac{dy}{dx} + x\frac{dy}{dx} = 2x - y \;\Rightarrow\; \frac{dy}{dx}(1+x) = 2x - y \;\Rightarrow\; \frac{dy}{dx} = \frac{2x-y}{1+x}.$ …

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10. KCET 2020Set A-11 markMCQ
    Q.If the curves $2x = y^2$ and $2xy = K$ intersect perpendicularly, then the value of $K^2$ is (A) 4 (B) $2\sqrt{2}$ (C) 2 (D) 8

    ›Reveal solutionSolution

    Differentiate each curve implicitly to get its slope at the common point, impose $m_1 m_2 = -1$, and solve for the intersection — then read off $K$.

    Step 1 — Slope of the parabola $2x = y^{2}$.

    Differentiate implicitly w.r.t. $x$:

    $2 = 2y\frac{dy}{dx} \quad\Longrightarrow\quad m_{1} = \frac{dy}{dx} = \frac{1}{y}.$

    Step 2 — Slope of the hyperbola $2xy = K$.

    Differentiate implicitly (product rule):

    $2\left(y + x\frac{dy}{dx}\right) = 0 \quad\Longrightarrow\quad m_{2} = \frac{dy}{dx} = -\frac{y}{x}.$

    Step 3 — Impose orthogonality (why: two curves cut at right angles $\iff$ the product of their tangent slopes at the common point is $-1$).

    $m_{1}m_{2} = -1 \;\Longrightarrow\; \left(\frac{1}{y}\right)\left(-\frac{y}{x}\right) = -1 \;\Longrightarrow\; -\frac{1}{x} = -1 \;\Longrightarrow\; x = 1.$

    Step 4 — Find $y$ at that point, from the parabola. …

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11. KCET 2021Set A-11 markMCQ
    Q.For constant $a$, $\frac{d}{dx} (x^x + x^a + a^x + a^a)$ is (A) $x^x (1 + \log x) + a x^{a-1}$ (B) $x^x (1 + \log x) + a x^{a-1} + a^x \log a$ (C) $x^x (1 + \log x) + a^a (1 + \log x)$ (D) $x^x (1 + \log x) + a^a (1 + \log a) + a x^{a-1}$

    ›Reveal solutionSolution

    Differentiate each term separately using the appropriate rule — power rule, exponential rule, and the special logarithmic differentiation for $x^x$. The derivative is $x^x(1+\log x) + a x^{a-1} + a^x \log a$, which matches option (B).

    The key is to recognise that $a$ is a constant, so $x^a$ and $a^x$ are standard forms, while $x^x$ requires logarithmic differentiation. The term $a^a$ is just a constant and differentiates to zero.

    1. Differentiate $x^x$ Write $y = x^x$. Take logs: $\log y = x \log x$. Differentiate both sides:

    $\frac{1}{y} \frac{dy}{dx} = \log x + x \cdot \frac{1}{x} = \log x + 1$

    So $\frac{dy}{dx} = y(1 + \log x) = x^x (1 + \log x)$.

    2. Differentiate $x^a$ Here $a$ is a constant exponent. Use the power rule:

    $\frac{d}{dx} x^a = a x^{a-1}$

    3. Differentiate $a^x$ Here $a$ is a constant base. Use the exponential rule:

    $\frac{d}{dx} a^x = a^x \log a$

    4. Differentiate $a^a$

       Since $a$ is constant, $a^a$ is a constant number. Its derivative is zero.

    5. Add all the derivatives …

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12. COMEDK 2021Set 2021-B1 markMCQ
    Q.The curve $y-e^{xy}+x=0$ has a vertical tangent at the point (A) (e, 0) (B) (1, 1) (C) (1, 0) (D) (0, 1)

    ›Reveal solutionSolution

    Vertical tangent occurs where $x\,e^{xy}=1$; the point $(1,0)$ satisfies both the curve and this condition.

    Curve: $y-e^{xy}+x=0$. Differentiate implicitly:

    $\dfrac{dy}{dx}-e^{xy}\!\left(y+x\dfrac{dy}{dx}\right)+1=0.$

    Collect terms:

    $\dfrac{dy}{dx}\big(1-x e^{xy}\big)=y e^{xy}-1\;\Rightarrow\;\dfrac{dy}{dx}=\dfrac{y e^{xy}-1}{1-x e^{xy}}.$

    A vertical tangent requires the denominator $=0$: $x e^{xy}=1$ (with numerator $\neq0$). …

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