Q.Derivative of $x^2$ w.r.t. $x^3$ is __________.

Showing the 12 most recent of 50 on this concept.

1. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If $f(x) = \sqrt{\cos^{-1} \sqrt{1-x^2}}$, then $f'\left(\dfrac{1}{2}\right) =$ (A) $\sqrt{\dfrac{2}{\pi}}$ (B) $\sqrt{\dfrac{\pi}{2}}$ (C) $-\sqrt{\dfrac{2}{\pi}}$ (D) $-\sqrt{\dfrac{\pi}{2}}$

   ›Reveal solutionSolution

   For $x\ge0$, $\cos^{-1}\sqrt{1-x^2}=\sin^{-1}x$, so $f(x)=\sqrt{\sin^{-1}x}$ and $f'\!\left(\tfrac12\right)=\sqrt{\tfrac{2}{\pi}}$ — option (A).

   Simplify. For $0\le x\le 1$, let $\theta=\cos^{-1}\sqrt{1-x^2}\in\big[0,\tfrac{\pi}{2}\big]$. Then $\cos^2\theta=1-x^2$, so $\sin\theta=x$ and $\theta=\sin^{-1}x$. Hence

   $f(x)=\sqrt{\sin^{-1}x}.$

   Differentiate.

   $f'(x)=\frac{1}{2\sqrt{\sin^{-1}x}}\cdot\frac{1}{\sqrt{1-x^2}}.$

   Evaluate at $x=\tfrac12$. $\sin^{-1}\tfrac12=\dfrac{\pi}{6}$ and $\sqrt{1-\tfrac14}=\dfrac{\sqrt3}{2}$: …

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2. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.If $2x^2 - 3xy + 4y^2 + 2x - 3y + 4 = 0$, then $\left(\dfrac{dy}{dx}\right)_{(3,2)} =$ (A) $-5$ (B) $\dfrac{5}{7}$ (C) $-2$ (D) $\dfrac{2}{7}$

   ›Reveal solutionSolution

   Implicit differentiation gives $\dfrac{dy}{dx}=\dfrac{-(4x-3y+2)}{-3x+8y-3}$, and at $(3,2)$ this evaluates to $\dfrac{-8}{4}=-2$. Answer: (C).

   Concept

   For a curve given implicitly, differentiate every term with respect to $x$ treating $y=y(x)$ (chain and product rules), then collect and isolate $\dfrac{dy}{dx}$.

   Solution

   1. Differentiate the relation $2x^2-3xy+4y^2+2x-3y+4=0$:

   $4x-3\!\left(y+x\frac{dy}{dx}\right)+8y\frac{dy}{dx}+2-3\frac{dy}{dx}=0.$

   2. Collect the derivative terms.

   $(-3x+8y-3)\frac{dy}{dx}+(4x-3y+2)=0,$

   $\frac{dy}{dx}=\frac{-(4x-3y+2)}{-3x+8y-3}.$

   3. Evaluate at $(3,2)$.

   $\text{Numerator: } -(4\cdot3-3\cdot2+2)=-(12-6+2)=-8,$

   $\text{Denominator: } -3\cdot3+8\cdot2-3=-9+16-3=4,$

   $\left.\frac{dy}{dx}\right|_{(3,2)}=\frac{-8}{4}=-2.$ …

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3. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.If $|x|$ is so small that $x^3$ and higher powers of $x$ can be neglected, then an approximate value of $\dfrac{1}{\sqrt{4-x(2+x)^3}}$ is (A) $\dfrac{1}{16} \left( 1 + \dfrac{13x}{8} + \dfrac{219}{128} x^2 \right)$ (B) $\dfrac{1}{16} \left( 1 + \dfrac{11x}{8} - \dfrac{165}{128} x^2 \right)$ (C) $\dfrac{1}{32} \left( 1 - \dfrac{11x}{8} + \dfrac{219}{128} x^2 \right)$ (D) $\dfrac{1}{16} \left( 1 - \dfrac{11x}{8} + \dfrac{171}{128} x^2 \right)$

   ›Reveal solutionSolution

   Expanding $\dfrac{1}{\sqrt{4-x}\,(2+x)^3}$ to order $x^2$ gives $\dfrac{1}{16}\!\left(1 - \dfrac{11x}{8} + \dfrac{171}{128}x^2\right)$ (option D).

   The expression is $\dfrac{1}{\sqrt{4-x}\,(2+x)^3}$ (value $\tfrac{1}{2\cdot 8}=\tfrac{1}{16}$ at $x=0$).

   First factor:

   $(4-x)^{-1/2} = \tfrac12\left(1-\tfrac{x}{4}\right)^{-1/2} = \tfrac12\!\left(1 + \tfrac{x}{8} + \tfrac{3x^2}{128}\right).$

   Second factor:

   $(2+x)^{-3} = \tfrac18\left(1+\tfrac{x}{2}\right)^{-3} = \tfrac18\!\left(1 - \tfrac{3x}{2} + \tfrac{3x^2}{2}\right).$

   Product ($\tfrac12\cdot\tfrac18 = \tfrac{1}{16}$):

   $\frac{1}{16}\left(1 + \tfrac{x}{8} + \tfrac{3x^2}{128}\right)\left(1 - \tfrac{3x}{2} + \tfrac{3x^2}{2}\right).$ …

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4. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
   Q.For the differential equation

   [!FORMULA] $\sqrt{\frac{d^2 y}{dx^2}} = 3 \sqrt{y \frac{dy}{dx} + x \sin \left( \frac{dy}{dx} \right)}^2$

   (A) Order is 2 and degree is 3 (B) Order is 3 and degree is 3 (C) Order is 3 and degree is 2 (D) Order is 2 and degree is not defined

   ›Reveal solutionSolution

   The highest derivative present is $\dfrac{d^2y}{dx^2}$, so the order is 2. Because $\dfrac{dy}{dx}$ appears inside $\sin(\cdot)$, the equation cannot be written as a polynomial in the derivatives, so the degree is not defined — option (D).

   The equation is

   $\sqrt{\frac{d^2y}{dx^2}}=3\sqrt{\,y\frac{dy}{dx}+x\sin\!\left(\frac{dy}{dx}\right)\,}^{\,2}.$

   Order. The order is the highest-order derivative that appears. Here that is $\dfrac{d^2y}{dx^2}$, so the order is $2$. …

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5. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.On differentiation if we get $f(x, y) \, dy - g(x, y) \, dx = 0$ from $2x^2 - 3xy + y^2 + x + 2y - 8 = 0$ then $\dfrac{g(2,2)}{f(1,1)}$ (A) $\dfrac{11}{7}$ (B) $-3$ (C) $-\dfrac{1}{3}$ (D) $7$

   ›Reveal solutionSolution

   The problem asks for the ratio $\frac{g(2,2)}{f(1,1)}$ where $f$ and $g$ come from rewriting the differential of the given implicit equation. We find $f$ and $g$ by implicit differentiation, then evaluate and compute the ratio, obtaining $-\frac{1}{3}$.

   We start with the equation

   $2x^2 - 3xy + y^2 + x + 2y - 8 = 0.$

   The phrase "On differentiation if we get $f(x, y) \, dy - g(x, y) \, dx = 0$" means we differentiate the equation implicitly and then rearrange terms so that all $dy$ terms are grouped together and all $dx$ terms are grouped together, with the form $f\,dy - g\,dx = 0$. Here $f$ and $g$ are functions of $x$ and $y$.

   Concept & Intuition:

   When we differentiate an implicit relation $F(x,y)=0$, we get $F_x\,dx + F_y\,dy = 0$. That can be rewritten as $F_y\,dy = -F_x\,dx$, or equivalently $F_y\,dy + F_x\,dx = 0$. But the problem wants the form $f\,dy - g\,dx = 0$, so we match: $f = F_y$ and $g = -F_x$. Then the ratio $\frac{g}{f}$ is just $-\frac{F_x}{F_y}$, which is the negative of the derivative $\frac{dy}{dx}$ (since $\frac{dy}{dx} = -\frac{F_x}{F_y}$). So we are essentially computing a ratio of partial derivatives at given points.

   Let’s do it step by step.

   1. Differentiate implicitly with respect to $x$, treating $y$ as a function of $x$:

   $\frac{d}{dx}(2x^2) - \frac{d}{dx}(3xy) + \frac{d}{dx}(y^2) + \frac{d}{dx}(x) + \frac{d}{dx}(2y) - \frac{d}{dx}(8) = 0.$

   This gives:

   $4x - 3\left( y + x\frac{dy}{dx} \right) + 2y\frac{dy}{dx} + 1 + 2\frac{dy}{dx} = 0.$

   2. Simplify:

   $4x - 3y - 3x\frac{dy}{dx} + 2y\frac{dy}{dx} + 1 + 2\frac{dy}{dx} = 0.$

   3. Collect terms with $\frac{dy}{dx}$:

   $(-3x + 2y + 2)\frac{dy}{dx} + (4x - 3y + 1) = 0.$

   4. Isolate $\frac{dy}{dx}$:

   $(-3x + 2y + 2)\frac{dy}{dx} = -(4x - 3y + 1).$

   So

   $\frac{dy}{dx} = -\frac{4x - 3y + 1}{-3x + 2y + 2}.$

   5. Rewrite in the form $f\,dy - g\,dx = 0$: Multiply both sides of the differentiated equation by $dx$:

   $(-3x + 2y + 2)\,dy + (4x - 3y + 1)\,dx = 0.$

   But the problem wants $f\,dy - g\,dx = 0$. So we need a minus sign before the $dx$ term. Write: …

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6. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
   Q.If $\tan h\,x = \dfrac{1}{2}$ then $\sin h\,2x - \sec h\,2x =$ (A) $\dfrac{29}{15}$ (B) $\dfrac{11}{15}$ (C) $3$ (D) $\dfrac{-13}{15}$

   ›Reveal solutionSolution

   Use the given $\tanh x = \frac12$ to find $\sinh 2x$ and $\sech 2x$ via hyperbolic identities, then subtract to get $\frac{29}{15}$.

   The core idea here is that hyperbolic functions obey identities very similar to trigonometric ones, but with sign differences. Given $\tanh x$, you can find $\sinh x$ and $\cosh x$ using the fundamental relation $\cosh^2 x - \sinh^2 x = 1$, then compute double-angle forms. The trick is to avoid solving for $x$ directly — work algebraically with the ratios.

   1. Find $\cosh x$ and $\sinh x$ from $\tanh x$. Let $\tanh x = \frac{\sinh x}{\cosh x} = \frac12$. So $\sinh x = \frac12 \cosh x$. Use $\cosh^2 x - \sinh^2 x = 1$:

   $\cosh^2 x - \left(\frac12 \cosh x\right)^2 = 1 \quad\Rightarrow\quad \cosh^2 x - \frac14 \cosh^2 x = \frac34 \cosh^2 x = 1.$

   Hence $\cosh^2 x = \frac43$, so $\cosh x = \frac{2}{\sqrt{3}}$ (positive, since $\cosh x > 0$ for all real $x$).

   Then $\sinh x = \frac12 \cdot \frac{2}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.

   2. Compute $\sinh 2x$. Using $\sinh 2x = 2 \sinh x \cosh x$:

   $\sinh 2x = 2 \cdot \frac{1}{\sqrt{3}} \cdot \frac{2}{\sqrt{3}} = \frac{4}{3}.$

   3. Compute $\cosh 2x$ and then $\sech 2x$. Use $\cosh 2x = \cosh^2 x + \sinh^2 x$ (note: plus sign, unlike $\cos 2x$): $$$$

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7. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.The function $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$ where $a > 0$ attains its local maximum and local minimum at $p$ and $q$ respectively. If $p^2 = q$ then $a =$ (A) $1$ (B) $2$ (C) $3$ (D) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   The key idea is to find the critical points of the cubic, express the condition $p^2 = q$ in terms of $a$, and solve the resulting equation. The answer is $a = 2$.

   We are given $f(x) = 2x^3 - 9a x^2 + 12a^2 x + 1$ with $a > 0$. The function attains a local maximum at $x = p$ and a local minimum at $x = q$, and we know $p^2 = q$. We need to find $a$.

   ---

   1. Find the critical points

   For a cubic with positive leading coefficient, the derivative is a quadratic. The critical points are the roots of $f'(x) = 0$.

   $f'(x) = 6x^2 - 18a x + 12a^2$

   Divide through by 6:

   $f'(x) = x^2 - 3a x + 2a^2$

   Factor:

   $x^2 - 3a x + 2a^2 = (x - a)(x - 2a)$

   So the critical points are $x = a$ and $x = 2a$.

   Since $a > 0$, we have $a < 2a$.

   ---

   2. Determine which is max and which is min

   For a cubic with positive leading coefficient, the derivative is a upward-opening parabola. The derivative changes sign from positive to negative at the first critical point (the smaller one), and from negative to positive at the second (larger) critical point.

   • At $x = a$ (the smaller root), $f'(x)$ goes from positive to negative → local maximum.
   • At $x = 2a$ (the larger root), $f'(x)$ goes from negative to positive → local minimum.

   Thus:

   $p = a \quad (\text{local max}), \qquad q = 2a \quad (\text{local min})$

   ---

   3. Apply the condition $p^2 = q$ …

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8. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
   Q.If the tangent at a point P on the curve $y = 4x^4 + x$ is perpendicular to the tangent to the same curve at (0, 0), then the point P is (A) $\left( \dfrac{-1}{2}, \dfrac{-1}{4} \right)$ (B) $\left( \dfrac{1}{2}, \dfrac{3}{4} \right)$ (C) $(1, 5)$ (D) $(-1, 3)$

   ›Reveal solutionSolution

   To find the point P, we first determine the slope of the tangent at (0, 0) using the derivative. Since the tangent at P is perpendicular to this, its slope must be the negative reciprocal. We then set the general derivative equal to this required slope to find the x-coordinate of P, and finally use the curve equation to find the y-coordinate. The point P is $\boxed{\left( \dfrac{-1}{2}, \dfrac{-1}{4} \right)}$.

   The slope of the tangent to a curve at any point is given by the derivative of the curve's equation with respect to $x$, evaluated at that point. This derivative, $\frac{dy}{dx}$, represents the instantaneous rate of change of $y$ with respect to $x$, which is precisely the slope of the line tangent to the curve.

   For two lines to be perpendicular, the product of their slopes must be $-1$. If one line has a slope $m_1$ and the other has a slope $m_2$, then $m_1 \cdot m_2 = -1$. This condition is key to solving the problem.

   Here's how we apply these concepts:

   1. Find the derivative of the curve: The given curve is $y = 4x^4 + x$. We differentiate $y$ with respect to $x$ to find the general expression for the slope of the tangent at any point $(x, y)$:

   $\frac{dy}{dx} = \frac{d}{dx}(4x^4 + x) = 16x^3 + 1$

   2. Calculate the slope of the tangent at (0, 0): Let $m_1$ be the slope of the tangent at the point $(0, 0)$. We substitute $x=0$ into the derivative:

   $m_1 = \left. \frac{dy}{dx} \right|_{x=0} = 16(0)^3 + 1 = 1$

   3. Determine the required slope for the tangent at point P: Let $m_2$ be the slope of the tangent at point P. We are given that the tangent at P is perpendicular to the tangent at (0, 0).

      For two perpendicular lines with slopes $m_1$ and $m_2$, we have $m_1 \cdot m_2 = -1$.

      Using this condition:

   $1 \cdot m_2 = -1$

   $m_2 = -1$

   So, the tangent at point P must have a slope of $-1$.
   

   4. Find the x-coordinate of point P:

   We know that the slope of the tangent at any point $(x, y)$ is $16x^3 + 1$. We set this equal to the required slope $m_2 = -1$:

   $16x^3 + 1 = -1$

   $16x^3 = -2$

   $x^3 = \frac{-2}{16}$

   $$ x^3 = \frac{-1}{8} $$ …
   

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9. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.The function $f(x) = 2x^3 - 9ax^2 + 12a^2 x + 1$ where $a > 0$ attains its local maximum and local minimum at $p$ and $q$ respectively. If $p^2 = q$ then $a =$ (A) $3$ (B) $2$ (C) $1$ (D) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   For a cubic with a positive leading coefficient, the derivative gives the stationary points; using the condition $p^2 = q$ and the relationship between the roots of the derivative yields $a = 2$.

   The key idea is that local maxima and minima of a polynomial occur where its derivative is zero. For a cubic, the derivative is a quadratic, so its two roots are exactly the $x$-coordinates of the local maximum and minimum. The problem gives a relation between these two roots, which lets us solve for $a$.

   1. Find the derivative and its roots. $f(x) = 2x^3 - 9a x^2 + 12a^2 x + 1$ $f'(x) = 6x^2 - 18a x + 12a^2$ Set $f'(x) = 0$ and divide through by 6:

   $x^2 - 3a x + 2a^2 = 0$

   This quadratic factors nicely:

   $(x - a)(x - 2a) = 0$

   So the stationary points are at $x = a$ and $x = 2a$.

   2. Identify which is the maximum and which is the minimum.

      Since $a > 0$, we have $a < 2a$. For a cubic with a positive leading coefficient ($2 > 0$), the graph rises from $-\infty$, reaches a local maximum, then a local minimum, then rises to $+\infty$. So the smaller root $x = a$ is the local maximum, and the larger root $x = 2a$ is the local minimum.

      Thus $p = a$ (local maximum) and $q = 2a$ (local minimum).

   3. Apply the given condition $p^2 = q$.

      Substitute $p = a$ and $q = 2a$:

   $a^2 = 2a$

   Since $a > 0$, we can divide by $a$:

   $a = 2$ …

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10. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
    Q.If $\int (\sqrt{\csc x + 1})\, dx = k \tan^{-1}(f(x)) + c$, then $\frac{1}{k} f\left(\frac{\pi}{6}\right) =$ (A) $\frac{1}{2}$ (B) $\frac{1}{4}$ (C) $-\frac{1}{4}$ (D) $-\frac{1}{2}$

    ›Reveal solutionSolution

    $\displaystyle\int\sqrt{\csc x + 1}\,dx = -2\tan^{-1}\!\left(\sqrt{\csc x - 1}\right) + c$, so $k=-2$, $f(x)=\sqrt{\csc x-1}$ and $\tfrac{1}{k}f\!\left(\tfrac{\pi}{6}\right) = -\tfrac{1}{2}$.

    Evaluating the integral.

    Write $\sqrt{\csc x + 1} = \sqrt{\dfrac{1+\sin x}{\sin x}}$. We claim

    $\int \sqrt{\csc x + 1}\,dx = -2\tan^{-1}\!\left(\sqrt{\csc x - 1}\right) + c.$

    Verification by differentiation.

    Let $g = \sqrt{\csc x - 1}$, so $g^2 = \csc x - 1$ and $1 + g^2 = \csc x$. Then

    $2g\,g' = -\csc x\cot x \quad\Rightarrow\quad g' = \frac{-\csc x\cot x}{2g}.$

    $\frac{d}{dx}\left[-2\tan^{-1} g\right] = \frac{-2g'}{1+g^2} = \frac{-2g'}{\csc x} = \frac{\cot x}{g} = \frac{\cos x}{\sin x\,\sqrt{\dfrac{1-\sin x}{\sin x}}} = \frac{\cos x}{\sqrt{\sin x}\,\sqrt{1-\sin x}}.$

    Since $\cos x = \sqrt{1-\sin^2 x} = \sqrt{1-\sin x}\,\sqrt{1+\sin x}$, this reduces to …

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11. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
    Q.If the tangent and the normal drawn to the curve $xy^2 + x^2y = 12$ at the point (1,3) meet the X-axis in T and N respectively, then TN = (A) $\dfrac{7}{5}$ (B) $\dfrac{45}{7}$ (C) $\dfrac{3\sqrt{274}}{7}$ (D) $\dfrac{274}{35}$

    ›Reveal solutionSolution

    The key idea is to find the tangent and normal lines at (1,3) on the curve, compute their x-intercepts T and N, then find the distance TN. The result is a rational number, and the correct option is (B).

    ---

    We are given the curve $xy^2 + x^2y = 12$ and the point $(1,3)$. The tangent and normal at this point meet the X-axis at points T and N respectively. We need the distance TN.

    Concept & Intuition

    The tangent line at a point is the best linear approximation to the curve; the normal is perpendicular to it. Their x-intercepts are found by setting $y=0$ in their equations. The distance between these intercepts is simply the absolute difference of their x-coordinates, since both lie on the X-axis. So the problem reduces to: find the slope of the tangent, write both line equations, find where each crosses the x-axis, and subtract.

    ---

    Step-by-step solution

    1. Implicit differentiation to find the slope of the tangent Differentiate $xy^2 + x^2y = 12$ with respect to $x$:

    $\frac{d}{dx}(xy^2) + \frac{d}{dx}(x^2y) = 0$

    Using the product rule:

    $y^2 + 2xy\frac{dy}{dx} + 2xy + x^2\frac{dy}{dx} = 0$

    Collect terms with $\frac{dy}{dx}$:

    $(2xy + x^2)\frac{dy}{dx} = - (y^2 + 2xy)$

    So

    $\frac{dy}{dx} = -\frac{y^2 + 2xy}{2xy + x^2}$

    2. Evaluate the slope at $(1,3)$ Substitute $x=1, y=3$:

    $m_T = -\frac{9 + 2(1)(3)}{2(1)(3) + 1} = -\frac{9+6}{6+1} = -\frac{15}{7}$

    So the tangent slope is $-\frac{15}{7}$.

    3. Equation of the tangent line Using point-slope form:

    $y - 3 = -\frac{15}{7}(x - 1)$

    To find its x-intercept T, set $y=0$:

    $-3 = -\frac{15}{7}(x_T - 1) \quad\Rightarrow\quad 3 = \frac{15}{7}(x_T - 1)$

    Multiply both sides by 7:

    $21 = 15(x_T - 1) \quad\Rightarrow\quad x_T - 1 = \frac{21}{15} = \frac{7}{5}$

    Hence

    $x_T = 1 + \frac{7}{5} = \frac{12}{5}$

    So T = $\left(\frac{12}{5}, 0\right)$.

    4. Equation of the normal line The slope of the normal is the negative reciprocal of the tangent slope:

    $m_N = \frac{7}{15}$

    Equation:

    $y - 3 = \frac{7}{15}(x - 1)$

    Set $y=0$ to find N:

    $-3 = \frac{7}{15}(x_N - 1) \quad\Rightarrow\quad -45 = 7(x_N - 1)$

    So

    $$$$

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12. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
    Q.If the tangent and the normal drawn to the curve $xy^2 + x^2y = 12$ at the point $(1,3)$ meet the X-axis in T and N respectively, then TN = (A) $\dfrac{3\sqrt{274}}{7}$ (B) $\dfrac{45}{7}$ (C) $\dfrac{274}{35}$ (D) $\dfrac{7}{5}$

    ›Reveal solutionSolution

    The problem asks for the distance between the X‑intercepts of the tangent and normal to the curve at (1,3). We find the slope by implicit differentiation, write the line equations, compute their X‑intercepts, and subtract. The result is $\frac{45}{7}$, so option (B) is correct.

    We start with the curve

    $xy^2 + x^2y = 12$

    and the point $(1,3)$. The tangent and normal at this point will each cross the X‑axis; we need the distance between those two intercepts.

    ---

    1. Find the slope of the tangent using implicit differentiation

    Differentiate both sides with respect to $x$:

    $\frac{d}{dx}(xy^2) + \frac{d}{dx}(x^2y) = 0$

    For the first term, use the product rule:

    $\frac{d}{dx}(xy^2) = 1\cdot y^2 + x\cdot 2y\frac{dy}{dx} = y^2 + 2xy\frac{dy}{dx}$

    For the second term:

    $\frac{d}{dx}(x^2y) = 2x\cdot y + x^2\cdot\frac{dy}{dx} = 2xy + x^2\frac{dy}{dx}$

    So the derivative equation is:

    $y^2 + 2xy\frac{dy}{dx} + 2xy + x^2\frac{dy}{dx} = 0$

    Collect the $\frac{dy}{dx}$ terms:

    $(2xy + x^2)\frac{dy}{dx} + (y^2 + 2xy) = 0$

    Thus

    $\frac{dy}{dx} = -\frac{y^2 + 2xy}{2xy + x^2}$

    At $(1,3)$:

    $m_T = -\frac{9 + 2\cdot 1\cdot 3}{2\cdot 1\cdot 3 + 1} = -\frac{9+6}{6+1} = -\frac{15}{7}$

    So the tangent slope is $-\frac{15}{7}$.

    ---

    2. Equation of the tangent and its X‑intercept

    Tangent line through $(1,3)$ with slope $-\frac{15}{7}$:

    $y - 3 = -\frac{15}{7}(x - 1)$

    Set $y=0$ to find the X‑intercept $T$:

    $0 - 3 = -\frac{15}{7}(x_T - 1) \quad\Rightarrow\quad -3 = -\frac{15}{7}(x_T - 1)$

    $3 = \frac{15}{7}(x_T - 1) \quad\Rightarrow\quad x_T - 1 = \frac{3\cdot 7}{15} = \frac{21}{15} = \frac{7}{5}$

    $x_T = 1 + \frac{7}{5} = \frac{12}{5}$

    So $T = \left(\frac{12}{5}, 0\right)$.

    ---

    3. Equation of the normal and its X‑intercept

    The normal slope is the negative reciprocal:

    $m_N = \frac{7}{15}$

    Normal line through $(1,3)$:

    $y - 3 = \frac{7}{15}(x - 1)$

    Set $y=0$ for the X‑intercept $N$:

    $0 - 3 = \frac{7}{15}(x_N - 1) \quad\Rightarrow\quad -3 = \frac{7}{15}(x_N - 1)$

    $x_N - 1 = -\frac{3\cdot 15}{7} = -\frac{45}{7}$

    $x_N = 1 - \frac{45}{7} = \frac{7 - 45}{7} = -\frac{38}{7}$

    So $N = \left(-\frac{38}{7}, 0\right)$.

    ---

    4. Distance TN

    Both points lie on the X‑axis, so the distance is simply the absolute difference of their x‑coordinates:

    $TN = \left|\frac{12}{5} - \left(-\frac{38}{7}\right)\right| = \frac{12}{5} + \frac{38}{7}$

    Find a common denominator (35):

    $\frac{12}{5} = \frac{84}{35}, \quad \frac{38}{7} = \frac{190}{35}$

    $TN = \frac{84 + 190}{35} = \frac{274}{35}$ …

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