Q.For the curve $\sqrt{x} + \sqrt{y} = 1$, $\dfrac{dy}{dx}$ at $\left(\dfrac{1}{4}, \dfrac{1}{4}\right)$ is __________.

Showing the 12 most recent of 50 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If $\tan(e^{3x}) = \cot(e^{2y})$, then at $x = 0$, $\dfrac{dy}{dx} =$ (A) $\dfrac{3}{2-\pi}$ (B) $\dfrac{2-\pi}{3}$ (C) $\dfrac{3}{\pi-2}$ (D) $\dfrac{\pi-2}{3}$

   ›Reveal solutionSolution

   Rewrite $\cot$ as a shifted $\tan$ to turn the equation into an algebraic (exponential) relation between $x$ and $y$, then implicitly differentiate and evaluate at $x=0$. Answer: $\dfrac{3}{2-\pi}$.

   Concept and Intuition

   $\tan\theta_1=\tan\theta_2$ implies $\theta_1=\theta_2+n\pi$ for integer $n$; taking $n=0$ (the principal relation intended here) converts the trig equation into a clean equation between the exponential expressions, which we can differentiate implicitly.

   Step-by-Step Solution

   1. Use the identity $\cot\theta = \tan\left(\dfrac{\pi}{2}-\theta\right)$ with $\theta=e^{2y}$: $\cot(e^{2y}) = \tan\left(\dfrac{\pi}{2}-e^{2y}\right)$.
   2. Given $\tan(e^{3x}) = \cot(e^{2y}) = \tan\left(\dfrac{\pi}{2}-e^{2y}\right)$, equate arguments (principal branch): $e^{3x} = \dfrac{\pi}{2}-e^{2y}$.
   3. Rearrange: $e^{3x}+e^{2y} = \dfrac{\pi}{2}$.
   4. Differentiate both sides w.r.t. $x$: $3e^{3x} + 2e^{2y}\dfrac{dy}{dx} = 0$.
   5. Solve: $\dfrac{dy}{dx} = -\dfrac{3e^{3x}}{2e^{2y}}$.
   6. At $x=0$: $e^{3x}=e^0=1$. From step 3, $1+e^{2y}=\dfrac{\pi}{2}\Rightarrow e^{2y}=\dfrac{\pi}{2}-1=\dfrac{\pi-2}{2}$. …

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2. 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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3. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If the normal drawn to the curve $y^4 = 16x^3$ at the point of intersection of this curve and the line $y = 2$ meets the X and Y axes at $A$ and $B$ respectively, then $OA + 3\,OB =$ (A) $6$ (B) $8$ (C) $16$ (D) $12$

   ›Reveal solutionSolution

   Finding the intersection point, then the normal line's slope and intercepts, gives $OA+3\,OB=12$.

   Concept and Intuition

   The normal to a curve at a point is the line perpendicular to the tangent there. We find the point of intersection with $y=2$, use implicit differentiation to get the tangent slope, take its negative reciprocal for the normal slope, then find where that normal line crosses each axis (its intercepts) to compute $OA$ and $OB$ (the distances from the origin to those intercepts).

   Step-by-Step Solution

   1. Find the intersection: substitute $y=2$ into $y^4=16x^3$: $16 = 16x^3 \Rightarrow x^3=1 \Rightarrow x=1$. So the point is $(1,2)$.
   2. Differentiate implicitly: $4y^3\dfrac{dy}{dx} = 48x^2 \Rightarrow \dfrac{dy}{dx} = \dfrac{12x^2}{y^3}$.
   3. At $(1,2)$: $\dfrac{dy}{dx} = \dfrac{12(1)}{8} = \dfrac32$. This is the tangent's slope.
   4. Normal's slope is the negative reciprocal: $m_{normal} = -\dfrac{2}{3}$.
   5. Normal line equation: $y - 2 = -\dfrac23(x-1)$. …

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4. 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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5. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.If $x^2+y^2+\sin y=4$, then the value of $\dfrac{d^2y}{dx^2}$ at the point $(-2,0)$ is (A) -34 (B) -32 (C) 34 (D) 32

   ›Reveal solutionSolution

   Implicit differentiation twice on $x^2+y^2+\sin y=4$ gives $y''=-34$ at $(-2,0)$.

   Concept and Intuition

   For an implicitly defined curve, differentiate the whole equation with respect to $x$ once to get $y'$ in terms of $x,y$, then differentiate that resulting equation again (product/chain rule carefully) to isolate $y''$, finally substituting the numeric point.

   Step-by-Step Solution

   1. Differentiate $x^2+y^2+\sin y=4$ w.r.t. $x$:

   $2x+2y\,y'+\cos y\,y'=0 \Rightarrow y'(2y+\cos y)=-2x \Rightarrow y'=\frac{-2x}{2y+\cos y}$

   2. At $(-2,0)$: $y'=\dfrac{-2(-2)}{2(0)+\cos0}=\dfrac{4}{1}=4$.
   3. Differentiate the equation $2x+2y y'+\cos y\, y'=0$ again w.r.t. $x$:

   $2+2(y')^2+2y\,y''-\sin y\,(y')^2+\cos y\,y''=0$

   $2+\left[2-\sin y\right](y')^2+(2y+\cos y)y''=0$

   4. At the point: $y=0$, $y'=4$, $\sin y=0$, $\cos y=1$, so $2y+\cos y=1$: …

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6. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.Let the normal drawn at a point $P$ on the curve $y^2-3x^2+y+10=0$ intersect the Y-axis at $\left(0,\dfrac{3}{2}\right)$. If $m$ is the slope of the tangent at $P$ to the curve, then $|m|=$ (A) 2 (B) 3 (C) 4 (D) 6

   ›Reveal solutionSolution

   Implicit differentiation gives the tangent slope in terms of $(x_0,y_0)$; using that the normal meets the $Y$-axis at $(0,3/2)$ pins down $y_0=1$, then the curve equation gives $x_0=\pm2$, so $|m|=4$.

   Concept and Intuition

   When a curve is given implicitly, differentiate term-by-term treating $y$ as a function of $x$ to get $dy/dx$ symbolically in terms of $x,y$. The normal line at a point has slope $-1/m$ where $m$ is the tangent slope; using where that normal crosses a known axis gives an equation linking $x_0,y_0$, which combined with the original curve equation pins down the point.

   Step-by-Step Solution

   1. Differentiate $y^2-3x^2+y+10=0$: $2y\,y' - 6x + y' = 0 \Rightarrow y'(2y+1)=6x \Rightarrow y' = \dfrac{6x}{2y+1}$.
   2. At $P=(x_0,y_0)$, tangent slope $m = \dfrac{6x_0}{2y_0+1}$, so normal slope $= -\dfrac{2y_0+1}{6x_0}$.
   3. Equation of the normal at $P$: $y - y_0 = -\dfrac{1}{m}(x-x_0)$. At $x=0$, $y=\dfrac{3}{2}$: $\dfrac{3}{2}-y_0 = -\dfrac{1}{m}(0-x_0) = \dfrac{x_0}{m}$.
   4. Since $m = \dfrac{6x_0}{2y_0+1}$, we get $\dfrac{x_0}{m} = \dfrac{2y_0+1}{6}$ (the $x_0$ cancels). …

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7. 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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8. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If $x^2y - xy^2 + x^3 - y^3 = 0$, then $\dfrac{dy}{dx}$ at the point $(1, 1)$ is (A) $1$ (B) $0$ (C) $-1$ (D) Does not exist

   ›Reveal solutionSolution

   Implicit differentiation of a symmetric cubic curve, evaluated at the point $(1,1)$.

   Concept and Intuition

   When a curve is given implicitly (not solved for $y$), differentiate every term with respect to $x$, treating $y$ as a function of $x$ and applying the product rule wherever $x$ and $y$ appear together. Collecting all the $y'$ terms on one side isolates the slope as a ratio of two expressions in $x,y$.

   Step-by-Step Solution

   1. Differentiate term by term: $\dfrac{d}{dx}(x^2y)=2xy+x^2y'$; $\dfrac{d}{dx}(xy^2)=y^2+2xyy'$; $\dfrac{d}{dx}(x^3)=3x^2$; $\dfrac{d}{dx}(y^3)=3y^2y'$.
   2. So $2xy+x^2y'-y^2-2xyy'+3x^2-3y^2y'=0$.
   3. Collect $y'$ terms: $y'(x^2-2xy-3y^2)=-(2xy-y^2+3x^2)$, i.e. $y'=\dfrac{-(2xy-y^2+3x^2)}{x^2-2xy-3y^2}=\dfrac{y^2-2xy-3x^2}{x^2-2xy-3y^2}$. …

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9. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.The length of the normal drawn to the curve $2x^3 + 2y^3 = 9xy$ at the point $(2, 1)$, is (A) $\dfrac{\sqrt{41}}{4}$ (B) $\dfrac{2}{3}\sqrt{41}$ (C) $\sqrt{5}$ (D) $\dfrac{2}{3}\sqrt{5}$

   ›Reveal solutionSolution

   Find the tangent slope by implicit differentiation, then apply the standard formula for the length of the normal at a point on a curve.

   Concept and Intuition

   The length of the normal segment (from the curve point down to where the normal line meets the $x$-axis) is $|y_1|\sqrt{1+m^2}$, where $m$ is the slope of the tangent at that point — this comes from the right triangle formed by the ordinate, the subnormal, and the normal itself.

   Step-by-Step Solution

   1. Differentiate $2x^3+2y^3=9xy$ implicitly: $6x^2+6y^2y'=9(y+xy')$, i.e. $6x^2+6y^2y'=9y+9xy'$.
   2. Collect: $y'(6y^2-9x)=9y-6x^2\Rightarrow y'=\dfrac{9y-6x^2}{6y^2-9x}$.
   3. At $(2,1)$: numerator $=9(1)-6(4)=9-24=-15$; denominator $=6(1)-9(2)=6-18=-12$.
   4. $y'=\dfrac{-15}{-12}=\dfrac{5}{4}=m$. …

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10. CA Foundation 2026Set may-20261 markMCQ
    Q.If $x^y = y^x$, then $\dfrac{dy}{dx} =$ ______. (A) $\dfrac{y(x \log y - y)}{x(y \log x - x)}$ (B) $\dfrac{y(x \log y + y)}{x(y \log x - x)}$ (C) $\dfrac{y(x \log y - y)}{x(y \log x + x)}$ (D) $\dfrac{x(x \log y - y)}{y(y \log x - x)}$

    ›Reveal solutionSolution

    Log-differentiate $x^y=y^x$: from $y\log x=x\log y$ you get $\dfrac{dy}{dx}=\dfrac{y(x\log y-y)}{x(y\log x-x)}$.

    Step 1 — Take logarithms of both sides

    $x^y=y^x\ \Rightarrow\ y\log x=x\log y$

    Step 2 — Differentiate implicitly with respect to $x$

    Apply the product rule to each side:

    $y'\log x+\frac{y}{x}=\log y+x\cdot\frac{y'}{y}$

    Step 3 — Collect the $y'$ terms

    $y'\log x-\frac{x\,y'}{y}=\log y-\frac{y}{x}$

    $y'\left(\log x-\frac{x}{y}\right)=\log y-\frac{y}{x}$

    Step 4 — Solve for $y'$ and tidy the fractions

    $y'=\frac{\log y-\dfrac{y}{x}}{\log x-\dfrac{x}{y}}=\frac{\dfrac{x\log y-y}{x}}{\dfrac{y\log x-x}{y}}=\frac{y\,(x\log y-y)}{x\,(y\log x-x)}$ …

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11. 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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12. 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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