Q.Find $\frac{dy}{dx}$ in the following: $x = 4t, y = \frac{4}{t}$

Showing the 12 most recent of 50 on this concept.

1. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If $x^2 + y^2 = t - \dfrac{1}{t}$ and $x^4 + y^4 = t^2 + \dfrac{1}{t^2}$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{y}{x}$ (B) $\dfrac{y^2}{x^2}$ (C) $\sqrt{\dfrac{y}{x}}$ (D) $-\dfrac{y}{x}$

   ›Reveal solutionSolution

   Eliminating the parameter $t$ between the two given equations produces the direct relation $x^2y^2=-1$ between $x$ and $y$, whose implicit derivative is $-y/x$.

   Concept and Intuition

   When $x$ and $y$ are both linked to a parameter $t$ through two equations, differentiating each with respect to $t$ separately (and dividing) works, but it is often faster — and here it is exact — to first eliminate $t$ algebraically to get a direct $x$–$y$ relation, then differentiate that implicitly in the ordinary way.

   Step-by-Step Solution

   1. Square the first equation: $(x^2+y^2)^2=\left(t-\dfrac1t\right)^2=t^2-2+\dfrac1{t^2}$, i.e. $x^4+2x^2y^2+y^4=t^2+\dfrac1{t^2}-2$.
   2. The second equation says $x^4+y^4=t^2+\dfrac1{t^2}$. Substitute this in: $\left(t^2+\dfrac1{t^2}\right)+2x^2y^2=t^2+\dfrac1{t^2}-2$.
   3. This forces $2x^2y^2=-2\Rightarrow x^2y^2=-1$ — a $t$-free relation directly linking $x$ and $y$. …

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2. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.If $x^2+y^2=t+\dfrac{1}{t}$ and $x^4+y^4=t^2+\dfrac{1}{t^2}$, then $x^3 y\dfrac{dy}{dx}=$ (A) -1 (B) 0 (C) 1 (D) 2

   ›Reveal solutionSolution

   The two given relations force $x^2y^2=1$ (i.e. $xy$ is constant), from which $x^3y\,dy/dx=-1$.

   Concept and Intuition

   Rather than solving for $x,y$ in terms of $t$ explicitly, combine the two given equations algebraically (square the first, subtract the second) to eliminate $t$ entirely and land on a simple constant-product relation between $x$ and $y$.

   Step-by-Step Solution

   1. Square the first relation: $(x^2+y^2)^2=\left(t+\dfrac1t\right)^2=t^2+2+\dfrac{1}{t^2}$, i.e.

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

   2. The second given relation is $x^4+y^4=t^2+\dfrac{1}{t^2}$.
   3. Subtract: $2x^2y^2 = \left(t^2+2+\dfrac1{t^2}\right)-\left(t^2+\dfrac1{t^2}\right)=2$, so $x^2y^2=1$.
   4. This means $xy=\pm1$, a constant independent of $t$. Differentiate $xy=\text{const}$ implicitly: …

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3. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.If $(a + \sqrt{2}\,b\cos x)(a - \sqrt{2}\,b\cos y) = a^2 - b^2$ where $a > b > 0$, then at $\left(\dfrac{\pi}{4}, \dfrac{\pi}{4}\right)$, $\dfrac{dy}{dx} =$ (A) $\dfrac{a + b}{a - b}$ (B) $\dfrac{a - b}{a + b}$ (C) $\dfrac{a - 2b}{a + 2b}$ (D) $\dfrac{2a + b}{2a - b}$

   ›Reveal solutionSolution

   Expand the product, simplify by dividing by the common factor $b$, then implicitly differentiate and evaluate at $x=y=\pi/4$. Answer: $\dfrac{a-b}{a+b}$.

   Concept and Intuition

   The given relation looks intimidating as a product, but expanding it cancels the $a^2$ on both sides (since the RHS is $a^2-b^2$) and leaves a much simpler equation relating $\cos x,\cos y$, and $\cos x\cos y$. From there it's routine implicit differentiation; the special evaluation point $x=y=\pi/4$ is chosen because $\sin$ and $\cos$ coincide there, which cancels neatly.

   Step-by-Step Solution

   1. Expand: $a^2 - a\sqrt2 b\cos y + a\sqrt2 b\cos x - 2b^2\cos x\cos y = a^2-b^2$.
   2. Cancel $a^2$ from both sides: $\sqrt2 ab(\cos x-\cos y) - 2b^2\cos x\cos y = -b^2$.
   3. Divide through by $b$ (nonzero): $\sqrt2 a(\cos x-\cos y) - 2b\cos x\cos y + b = 0$.
   4. Differentiate implicitly w.r.t. $x$ (treat $y=y(x)$):

   $\sqrt2 a\left(-\sin x + \sin y\cdot y'\right) - 2b\left(-\sin x\cos y - \cos x\sin y\cdot y'\right) = 0.$

   5. Group $y'$ terms: $y'\left(\sqrt2 a\sin y + 2b\cos x\sin y\right) = \sqrt2 a\sin x - 2b\sin x\cos y$.
   6. So $y' = \dfrac{\sin x(\sqrt2 a - 2b\cos y)}{\sin y(\sqrt2 a + 2b\cos x)}$.
   7. At $x=y=\dfrac\pi4$: $\sin x=\sin y=\cos x=\cos y=\dfrac1{\sqrt2}$. Substitute: …

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4. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.If $3\sin xy + 4\cos xy = 5$, then $\dfrac{dy}{dx}$ is equal to ____ (A) $\dfrac{3\sin xy + 4\cos xy}{3\cos xy - 4\sin xy}$ (B) $\dfrac{3\cos xy + 4\sin xy}{4\cos xy - 3\sin xy}$ (C) $\dfrac{-y}{x}$ (D) $\dfrac{x}{y}$

   ›Reveal solutionSolution

   Since $3\sin\theta+4\cos\theta$ has maximum value exactly $5$ (as $\sqrt{3^2+4^2}=5$), equating it to $5$ forces $xy$ to be a fixed constant, so implicit differentiation of $xy=c$ gives $dy/dx=-y/x$.

   Concept and Intuition

   $a\sin\theta+b\cos\theta$ always has amplitude $\sqrt{a^2+b^2}$ — here $\sqrt{9+16}=5$. So the equation $3\sin(xy)+4\cos(xy)=5$ isn't a "generic" implicit curve; it can only be satisfied when the expression sits exactly at its maximum, which happens at one specific angle. That pins $xy$ to a single constant value, turning a trigonometric-looking implicit relation into the much simpler $xy=\text{const}$.

   Step-by-Step Solution

   1. Note $3^2+4^2=25=5^2$, so $3\sin\theta+4\cos\theta$ has maximum value $5$, attained only when $\theta$ equals the specific angle $\phi=\tan^{-1}(3/4)$ (mod $2\pi$).
   2. The given equation demands $3\sin(xy)+4\cos(xy)=5$, i.e. the maximum — so $xy=\phi$ is fixed, a constant independent of which point on the curve we pick. …

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5. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.If $x^2 \tan^{-1}\dfrac{y}{x} - y^2 \tan^{-1}\dfrac{x}{y} = k$, then $\left(\dfrac{dy}{dx}\right)_{(1,1)} =$ (A) $0$ (B) $\pi/4$ (C) $1$ (D) $\pi/2$

   ›Reveal solutionSolution

   Implicit differentiation of $x^2\tan^{-1}(y/x)-y^2\tan^{-1}(x/y)=k$ evaluated at $(1,1)$ gives $dy/dx=1$.

   Concept and Intuition

   Differentiate both terms using product and chain rules, being careful with the derivatives of $\tan^{-1}(y/x)$ and $\tan^{-1}(x/y)$ with respect to $x$ (treating $y$ as a function of $x$).

   Step-by-Step Solution

   1. $\dfrac{d}{dx}\left[x^2\tan^{-1}\dfrac{y}{x}\right] = 2x\tan^{-1}\dfrac{y}{x} + x^2\cdot\dfrac{y'x-y}{x^2+y^2}$.
   2. $\dfrac{d}{dx}\left[y^2\tan^{-1}\dfrac{x}{y}\right] = 2yy'\tan^{-1}\dfrac{x}{y} + y^2\cdot\dfrac{y-xy'}{x^2+y^2}$.
   3. Setting derivative of LHS $=0$ and evaluating at $(1,1)$, where $\tan^{-1}(1)=\pi/4$ for both terms: $2\cdot\dfrac\pi4 + \dfrac{y'-1}{2} - 2y'\cdot\dfrac\pi4 - \dfrac{1-y'}{2} = 0$. …

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6. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If $\mathrm{Tan}^{-1}x^2 + \mathrm{Tan}^{-1}y^2 = \dfrac{\pi}{2}$, then $\left(\dfrac{dy}{dx}\right)_{(-1,2)} =$ (A) $0$ (B) $1$ (C) $\dfrac{1}{2}$ (D) $-\dfrac{1}{2}$

   ›Reveal solutionSolution

   Reducing to $y^2=x^{-2}$ gives $\dfrac{dy}{dx}=-\dfrac{1}{x^3 y}$, which at $(-1,2)$ equals $\dfrac12$.

   Concept and Intuition

   If $\mathrm{Tan}^{-1}a+\mathrm{Tan}^{-1}b=\dfrac{\pi}{2}$ with $a,b>0$, then $\mathrm{Tan}^{-1}b=\dfrac{\pi}{2}-\mathrm{Tan}^{-1}a=\mathrm{Cot}^{-1}a$, so $b=\dfrac{1}{a}$. Applying this to $a=x^2,\ b=y^2$ collapses the relation into an algebraic one.

   Step-by-Step Solution

   1. From $\mathrm{Tan}^{-1}x^2+\mathrm{Tan}^{-1}y^2=\dfrac{\pi}{2}$ we get $y^2=\dfrac{1}{x^2}=x^{-2}$.
   2. Differentiate: $2y\,\dfrac{dy}{dx}=-2x^{-3}$.
   3. Hence $\dfrac{dy}{dx}=-\dfrac{1}{x^3 y}$. …

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7. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.If $\dfrac{y}{x}\cos^4\alpha + \dfrac{x}{y}\sin^4\alpha = 2\sin^2\alpha \cdot \cos^2\alpha$, then $\dfrac{dy}{dx} =$ (A) $\sin^3\alpha \cos\alpha$ (B) $\sin^2\alpha \cos^2\alpha$ (C) $\dfrac{\sin^2\alpha}{\cos^2\alpha}$ (D) $\sin\alpha \cos^3\alpha$

   ›Reveal solutionSolution

   The given relation is secretly a perfect square in disguise; it forces $y=x\tan^2\alpha$, so $dy/dx=\tan^2\alpha$.

   Concept and Intuition

   Rather than differentiating implicitly right away, it pays to recognise the algebraic structure first. Multiplying by $xy$ converts the equation into a quadratic in $x$ and $y$ that factors as a perfect square, revealing $y/x$ is actually a constant (independent of $x$), which makes the derivative trivial.

   Step-by-Step Solution

   1. Start from $\dfrac{y}{x}\cos^4\alpha+\dfrac{x}{y}\sin^4\alpha = 2\sin^2\alpha\cos^2\alpha$.
   2. Multiply both sides by $xy$: $y^2\cos^4\alpha + x^2\sin^4\alpha = 2xy\sin^2\alpha\cos^2\alpha$.
   3. Rearrange: $y^2\cos^4\alpha - 2xy\sin^2\alpha\cos^2\alpha + x^2\sin^4\alpha = 0$.
   4. This is $(y\cos^2\alpha - x\sin^2\alpha)^2 = 0$, so $y\cos^2\alpha = x\sin^2\alpha$, i.e. $y = x\tan^2\alpha$. …

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8. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.The length of the tangent drawn at the point $p\left(\dfrac{\pi}{4}\right)$ on the curve $x^{2/3}+y^{2/3}=2^{2/3}$ is (A) $\dfrac{2}{3}$ (B) 1 (C) $\dfrac{4}{3}$ (D) 2

   ›Reveal solutionSolution

   A length-of-tangent computation on the astroid at its symmetric point $\theta=\pi/4$; the length comes out to exactly $1$.

   Concept and Intuition

   The astroid $x^{2/3}+y^{2/3}=a^{2/3}$ can be parametrized as $x=a\cos^3\theta,\ y=a\sin^3\theta$, which automatically satisfies the equation. "The point $P(\pi/4)$" refers to this parameter value. The length of the tangent line segment (from the point of tangency to where it meets the $x$-axis) is given by the standard formula $\left|\dfrac{y\sqrt{1+y'^2}}{y'}\right|$.

   Step-by-Step Solution

   1. Here $a=2$ (since $2^{2/3}=a^{2/3}$), so $x=2\cos^3\theta$, $y=2\sin^3\theta$.
   2. At $\theta=\pi/4$: $\cos(\pi/4)=\sin(\pi/4)=1/\sqrt2$, so $x=y=2\left(\dfrac{1}{\sqrt2}\right)^3=2\cdot\dfrac{1}{2\sqrt2}=\dfrac{1}{\sqrt2}$.
   3. Differentiate $x^{2/3}+y^{2/3}=a^{2/3}$ implicitly: $\dfrac23x^{-1/3}+\dfrac23y^{-1/3}y'=0\Rightarrow y'=-\left(\dfrac{y}{x}\right)^{1/3}$.
   4. Since $x=y$ here, $y'=-1$. …

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9. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If $\sqrt{x-xy}+\sqrt{y-xy}=1$, then $\dfrac{dy}{dx}=$ (A) $-\sqrt{\dfrac{y-y^2}{x-x^2}}$ (B) $-\sqrt{\dfrac{1-y^2}{1-x^2}}$ (C) $-\sqrt{\dfrac{1-y}{1-x}}$ (D) $-\sqrt{\dfrac{x-y}{x+y}}$

   ›Reveal solutionSolution

   Squaring the constraint reveals that it forces $x+y=1$ identically, so $\dfrac{dy}{dx}=-1$ throughout; matching this against the options singles out $-\sqrt{\dfrac{y-y^2}{x-x^2}}$, since it equals $-1$ for every point satisfying $y=1-x$.

   Concept and Intuition

   Rather than blindly grinding through implicit differentiation of two square roots, it pays to first understand the curve itself. Squaring $\sqrt{x-xy}+\sqrt{y-xy}=1$ carefully (using $s=x+y,\,p=xy$) collapses to a perfect square equalling zero, revealing that the relation is nothing but the straight line $x+y=1$. Once we know that, $\dfrac{dy}{dx}=-1$ is immediate, and we just need to find which option reduces to $-1$ on this line.

   Step-by-Step Solution

   1. Square the given equation:

   $x(1-y)+y(1-x)+2\sqrt{xy(1-x)(1-y)} = 1$

   $x+y-2xy+2\sqrt{xy(1-x)(1-y)} = 1$

   2. Let $s=x+y,\ p=xy$. Then:

   $2\sqrt{p(1-x)(1-y)} = 1-s+2p$

   Note $(1-x)(1-y)=1-s+p$. Squaring again:

   $4p(1-s+p) = (1-s+2p)^2$

   Let $q=1-s$. Expanding both sides: LHS $=4pq+4p^2$; RHS $=q^2+4pq+4p^2$. So $0=q^2$, i.e. $q=0$, i.e. $s=1$.

   3. Hence $x+y=1$ is forced — the given relation is the line $y=1-x$ (restricted to the domain where the square roots are real, $0\le x,y\le1$).

   4. Differentiating $y=1-x$ directly: $\dfrac{dy}{dx}=-1$.

   5. Check which option gives $-1$ identically along $y=1-x$:

   • (A): $y-y^2=y(1-y)$. Substituting $y=1-x$: $y(1-y)=(1-x)\cdot x=x-x^2$. So the ratio is exactly $1$, and $-\sqrt1=-1$ for every $x$. ✓ …

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10. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
    Q.If $y^{y^{y^{\cdot^{\cdot^{\cdot^{\infty}}}}}} = \log\{x + \log\{x + \cdots\}\}$, then $\dfrac{dy}{dx}$ at $x = e^2 - 2$, $y = \sqrt{2}$ equals _____ (A) $\dfrac{\log 2}{2\sqrt{2}(e^2 - 1)}$ (B) $\dfrac{1 - \log 2}{2\sqrt{2}(e^2 - 1)}$ (C) $\dfrac{\sqrt{2}(1 - \log 2)}{e^2 - 1}$ (D) $\dfrac{\log 2}{\sqrt{2}(e^2 - 1)}$

    ›Reveal solutionSolution

    Both sides define the same implicit quantity $u$ via a self-referential equation; differentiate each side's defining equation implicitly and combine using the chain rule. The answer is (B).

    Concept and Intuition

    The infinite power tower $y^{y^{y^{\cdots}}}=u$ satisfies the self-consistency equation $u=y^u$ (the tower "regenerates" itself). Likewise the infinite nested logarithm $\log\{x+\log\{x+\cdots\}\}=u$ satisfies $u=\log(x+u)$. Since the problem states these two quantities are equal (both equal to the same $u$), $u$ is implicitly a common function linking $x$ and $y$; differentiating each defining relation gives $du/dy$ and $du/dx$, and the chain rule combines them into $dy/dx$.

    Step-by-Step Solution

    1. Verify $u=2$ at the given point. Nested log: $u=\log(x+u)$ at $x=e^2-2$: try $u=2$: $\log(e^2-2+2)=\log(e^2)=2$ ✓. Tower: $u=y^u$ at $y=\sqrt2$: try $u=2$: $(\sqrt2)^2=2$ ✓. Both consistent with $u=2$.
    2. Differentiate the tower relation $u=y^u$ w.r.t. $y$. Take $\log$: $\log u = u\log y$. Differentiate: $\dfrac1u\dfrac{du}{dy} = \dfrac{du}{dy}\log y + \dfrac uy$ $\Rightarrow \left(\dfrac1u-\log y\right)\dfrac{du}{dy}=\dfrac uy \Rightarrow \dfrac{du}{dy}=\dfrac{u^2}{y(1-u\log y)}$. At $y=\sqrt2,\ u=2$: $\dfrac{du}{dy}=\dfrac{4}{\sqrt2(1-2\log\sqrt2)}=\dfrac{4}{\sqrt2(1-\ln2)}=\dfrac{2\sqrt2}{1-\ln2}$. …

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11. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
    Q.If the tangent to the curve $xy + ax + by = 0$ at (1,1) makes an angle $\tan^{-1}2$ with X-axis, then $\dfrac{ab}{a+b} =$ (A) $1$ (B) $2$ (C) $3$ (D) $4$

    ›Reveal solutionSolution

    Use the point lying on the curve plus the given slope at that point to get two linear equations in $a,b$. Answer: $\dfrac{ab}{a+b}=2$.

    Concept and Intuition

    A point on an implicitly-defined curve gives one equation relating the curve's parameters; the slope condition (via implicit differentiation) gives a second. Two equations, two unknowns.

    Step-by-Step Solution

    1. Since $(1,1)$ lies on $xy+ax+by=0$: $1+a+b=0 \Rightarrow a+b=-1$.
    2. Differentiate implicitly: $y+xy'+a+by'=0 \Rightarrow y'(x+b) = -(y+a) \Rightarrow y'=-\dfrac{y+a}{x+b}$.
    3. At $(1,1)$, slope $=\tan^{-1}2 \Rightarrow y'=2$: $-\dfrac{1+a}{1+b}=2 \Rightarrow -(1+a)=2(1+b) \Rightarrow -1-a=2+2b \Rightarrow a+2b=-3$.
    4. Subtract $a+b=-1$ from $a+2b=-3$: $b=-2$. Then $a=-1-b=-1-(-2)=1$. …

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12. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
    Q.The length of the tangent drawn at the point $P(1, 3\sqrt{3})$ on the curve $x^{2/3} + y^{2/3} = 4$ is (A) 4 (B) 6 (C) 12 (D) 8

    ›Reveal solutionSolution

    This tests the "length of tangent" formula (segment of the tangent line between the point of contact and the $x$-axis) applied to a point on an astroid. Answer: length $=6$.

    Concept and Intuition

    For a curve $y=f(x)$, the tangent line at $(x_1,y_1)$ with slope $m$ meets the $x$-axis at $\left(x_1-\dfrac{y_1}{m},0\right)$. The distance from $(x_1,y_1)$ to that point is called the length of the tangent, and it works out to

    $L=|y_1|\frac{\sqrt{1+m^2}}{|m|}.$

    So we just need $y_1$ and the slope $m=\left.\dfrac{dy}{dx}\right|_P$ from the implicit equation of the astroid.

    Step-by-Step Solution

    1. The curve is $x^{2/3}+y^{2/3}=4$. Check $P(1,3\sqrt3)$: $1^{2/3}=1$ and $(3\sqrt3)^{2/3}=(3^{3/2})^{2/3}=3$, so $1+3=4$ — $P$ lies on the curve.
    2. Differentiate implicitly: $\dfrac{2}{3}x^{-1/3}+\dfrac{2}{3}y^{-1/3}\dfrac{dy}{dx}=0\Rightarrow \dfrac{dy}{dx}=-\left(\dfrac{y}{x}\right)^{1/3}$. …

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