Q.If $y = (\cos x)^{(\cos x)^{(\cos x)^{\cdots\infty}}}$, show that $\dfrac{dy}{dx} = \dfrac{y^2 \tan x}{y \log \cos x - 1}$.

Showing the 12 most recent of 17 on this concept.

1. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.Derivative of $(\sin x)^x$ with respect to $x^{(\sin x)}$ is (A) $\dfrac{(\sin x)^{x-1} \left[ (\sin x) \log(\sin x) + x \cos x \right]}{x^{(\sin x)} \left[ x \cos x (\log x) + \sin x \right]}$ (B) $\dfrac{(\sin x)^x \left[ (\sin x) (\log(\sin x)) + x \cos x \right]}{x^{(\sin x)} \left[ x \cos x (\log x) + \sin x \right]}$ (C) $\dfrac{x^{\sin x - 1} \left[ x \cos x (\log x) + \sin x \right]}{(\sin x)^{x-1} \left[ (\sin x) \log(\sin x) + x \cos x \right]}$ (D) $\dfrac{x^{\sin x} \left[ x \cos x (\log x) + \sin x \right]}{(\sin x)^x \left[ (\sin x) \log(\sin x) + x \cos x \right]}$

   ›Reveal solutionSolution

   Differentiate both $u=(\sin x)^x$ and $v=x^{\sin x}$ logarithmically, then form $\frac{du/dx}{dv/dx}$. The numerator is $(\sin x)^{x-1}[\sin x\log(\sin x)+x\cos x]$ and the denominator carries $[x\cos x\log x+\sin x]$ — option (A).

   The concept first. Two ideas combine.

   1. Derivative of one function w.r.t. another. By the chain rule, $\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx}$. So we never need to eliminate $x$; we just differentiate each separately and divide.
   2. Logarithmic differentiation. Neither the power rule ($x^n$, constant exponent) nor the exponential rule ($a^x$, constant base) applies when both base and exponent depend on $x$. Taking $\log$ first turns the exponent into a product, which the product rule can handle.

   Step 1 — Differentiate $u = (\sin x)^{x}$.

   $\log u = x\,\log(\sin x)$

   Differentiate both sides:

   $\frac{1}{u}\frac{du}{dx} = \log(\sin x) + x\cdot\frac{\cos x}{\sin x}$

   $\frac{du}{dx} = (\sin x)^{x}\left[\log(\sin x) + \frac{x\cos x}{\sin x}\right] = (\sin x)^{x-1}\Big[\sin x\,\log(\sin x) + x\cos x\Big]$

   (the last step just takes one factor of $\sin x$ out of the bracket).

   Step 2 — Differentiate $v = x^{\sin x}$.

   $\log v = \sin x \,\log x$

   $\frac{1}{v}\frac{dv}{dx} = \cos x\,\log x + \frac{\sin x}{x}$

   $\frac{dv}{dx} = x^{\sin x}\left[\cos x \log x + \frac{\sin x}{x}\right] = x^{\sin x}\cdot\frac{1}{x}\Big[x\cos x\,\log x + \sin x\Big]$

   Step 3 — Divide. …

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2. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.A function $f : \mathbb{R} \to \mathbb{R}$ is such that $yf(x+y) + \cos my = 1 + yf(x)$. If $m = 2$, then $f'(x) =$ (A) $-2 \sin 2xy$ (B) $4x$ (C) $\dfrac{2 \sin 2xy}{y}$ (D) $2x^2$

   ›Reveal solutionSolution

   Rearranging the relation gives $\dfrac{f(x+y)-f(x)}{y} = \dfrac{1-\cos(mxy)}{y^{2}}$; letting $y\to0$ and using $1-\cos\theta \to \theta^{2}/2$ yields $f'(x) = \dfrac{m^{2}x^{2}}{2} = \mathbf{2x^{2}}$ for $m=2$ — option (D).

   The concept first. The derivative is defined by

   $f'(x) = \lim_{y \to 0} \frac{f(x+y) - f(x)}{y}$

   So whenever a problem hands you a relation connecting $f(x+y)$ and $f(x)$, the strategy is always the same: isolate $f(x+y) - f(x)$, divide by $y$, and take the limit. The answer must be a function of $x$ alone — $y$ is the vanishing increment, so any option still containing $y$ (like (A) and (C)) cannot be a derivative at all.

   Step 1 — Isolate the difference.

   $y f(x+y) + \cos(mxy) = 1 + y f(x)$

   $\Rightarrow\; y f(x+y) - y f(x) = 1 - \cos(mxy)$

   $\Rightarrow\; y\big[f(x+y) - f(x)\big] = 1 - \cos(mxy)$

   Step 2 — Build the difference quotient. Divide both sides by $y^2$ (valid for $y \ne 0$):

   $\frac{f(x+y) - f(x)}{y} = \frac{1 - \cos(mxy)}{y^{2}}$

   Step 3 — Take the limit $y \to 0$.

   Use the standard limit $1 - \cos\theta = 2\sin^{2}\!\left(\tfrac{\theta}{2}\right)$, so for small $\theta$, $1-\cos\theta \approx \tfrac{\theta^{2}}{2}$. With $\theta = mxy$: …

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3. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.If $(a + bx)e^{\frac{y}{x}} = x$, then $\frac{d^2 y}{dx^2} =$ (A) $\frac{1}{x^3} (xy' + y^2)^2$ (B) $\frac{1}{x^3} (xy' + y^2)$ (C) $\frac{1}{x^3} (xy' - y)$ (D) $\frac{1}{x^3} (xy' - y)^2$

   ›Reveal solutionSolution

   By simplifying the given equation using logarithms and then applying implicit differentiation twice, we find that the second derivative $\frac{d^2 y}{dx^2}$ is $\boxed{\frac{1}{x^3} (xy' - y)^2}$.

   The problem asks us to find the second derivative $\frac{d^2 y}{dx^2}$ from the given implicit relation $(a + bx)e^{\frac{y}{x}} = x$. The presence of the exponential term $e^{\frac{y}{x}}$ suggests that taking the natural logarithm might simplify the expression, making differentiation easier. The options provided involve the term $(xy' - y)$, which is a strong hint that we should try to express our derivatives in terms of this quantity.

   Here's a step-by-step approach:

   1. Simplify the given equation:

      The initial equation is $(a + bx)e^{\frac{y}{x}} = x$.

      To simplify, first isolate the exponential term:

      $e^{\frac{y}{x}} = \frac{x}{a + bx}$

      Now, take the natural logarithm on both sides. This brings the exponent down, making the equation linear in $\frac{y}{x}$:

      $\log\left(e^{\frac{y}{x}}\right) = \log\left(\frac{x}{a + bx}\right)$

      Using the logarithm property $\log(A/B) = \log A - \log B$:

      $\frac{y}{x} = \log x - \log(a + bx)$

      Multiplying by $x$ gives us an explicit expression for $y$:

      $y = x(\log x - \log(a + bx))$

   2. Find the first derivative, $y'$:

      We differentiate $y = x(\log x - \log(a + bx))$ with respect to $x$. We will use the product rule, $(uv)' = u'v + uv'$.

      Let $u = x$, so $u' = 1$.

      Let $v = \log x - \log(a + bx)$. To find $v'$, we differentiate term by term:

      $\frac{d}{dx}(\log x) = \frac{1}{x}$

      $\frac{d}{dx}(\log(a + bx)) = \frac{1}{a + bx} \cdot \frac{d}{dx}(a + bx) = \frac{b}{a + bx}$

      So, $v' = \frac{1}{x} - \frac{b}{a + bx}$.

      Applying the product rule for $y'$:

      $y' = (1) \cdot (\log x - \log(a + bx)) + x \cdot \left(\frac{1}{x} - \frac{b}{a + bx}\right)$

      $y' = (\log x - \log(a + bx)) + 1 - \frac{bx}{a + bx}$

      From Step 1, we know that $\log x - \log(a + bx) = \frac{y}{x}$. Substitute this back into the expression for $y'$:

      $y' = \frac{y}{x} + 1 - \frac{bx}{a + bx}$

      Combine the constant and fractional terms:

      $y' = \frac{y}{x} + \frac{(a + bx) - bx}{a + bx}$

      $y' = \frac{y}{x} + \frac{a}{a + bx}$

   3. Express $xy' - y$:

      The options involve the term $(xy' - y)$. Let's rearrange our expression for $y'$ from Step 2 to find this:

      $y' - \frac{y}{x} = \frac{a}{a + bx}$

      Multiply the entire equation by $x$:

      $x\left(y' - \frac{y}{x}\right) = x\left(\frac{a}{a + bx}\right)$

      $xy' - y = \frac{ax}{a + bx}$

      This is a key intermediate result.

   4. Find the second derivative, $y''$:

      Now we differentiate $y' = \frac{y}{x} + \frac{a}{a + bx}$ with respect to $x$ to find $y''$.

      $y'' = \frac{d}{dx}\left(\frac{y}{x}\right) + \frac{d}{dx}\left(\frac{a}{a + bx}\right)$

      For the first term, $\frac{d}{dx}\left(\frac{y}{x}\right)$, use the quotient rule $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$:

      $\frac{d}{dx}\left(\frac{y}{x}\right) = \frac{y' \cdot x - y \cdot 1}{x^2} = \frac{xy' - y}{x^2}$

      For the second term, $\frac{d}{dx}\left(\frac{a}{a + bx}\right)$, treat it as $a(a + bx)^{-1}$ and use the chain rule:

      $\frac{d}{dx}\left(a(a + bx)^{-1}\right) = a \cdot (-1)(a + bx)^{-2} \cdot \frac{d}{dx}(a + bx)$

      $= -a(a + bx)^{-2} \cdot b = -\frac{ab}{(a + bx)^2}$

      Combining these, we get $y''$: …

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4. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.If $x = \sin 2\theta \cos 3\theta$, $y = \sin 3\theta \cos 2\theta$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{2\cos 5\theta + \sin 3\theta \sin 2\theta}{2\cos 5\theta - \cos 3\theta \cos 2\theta}$ (B) $\dfrac{2\cos 5\theta - \sin 3\theta \sin 2\theta}{2\cos 5\theta + \cos 3\theta \cos 2\theta}$ (C) $\dfrac{2\cos 5\theta + \cos 3\theta \cos 2\theta}{2\cos 5\theta - \sin 3\theta \sin 2\theta}$ (D) $\dfrac{2\cos 5\theta - \sin 3\theta \sin 2\theta}{2\cos 5\theta - \cos 3\theta \cos 2\theta}$

   ›Reveal solutionSolution

   $\dfrac{dy}{dx}=\dfrac{2\cos5\theta+\cos3\theta\cos2\theta}{2\cos5\theta-\sin3\theta\sin2\theta}$.

   Differentiate each with respect to $\theta$:

   $\frac{dx}{d\theta}=2\cos2\theta\cos3\theta-3\sin2\theta\sin3\theta,$

   $\frac{dy}{d\theta}=3\cos3\theta\cos2\theta-2\sin3\theta\sin2\theta.$

   Using $\cos5\theta=\cos2\theta\cos3\theta-\sin2\theta\sin3\theta$, write each derivative around $\cos5\theta$:

   $\frac{dx}{d\theta}=2(\cos2\theta\cos3\theta-\sin2\theta\sin3\theta)-\sin3\theta\sin2\theta=2\cos5\theta-\sin3\theta\sin2\theta,$ …

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5. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.If $x = \sin 2\theta \cos 3\theta$, $y = \sin 3\theta \cos 2\theta$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{2\cos 5\theta + \cos 3\theta \cos 2\theta}{2\cos 5\theta - \sin 3\theta \sin 2\theta}$ (B) $\dfrac{2\cos 5\theta + \sin 3\theta \sin 2\theta}{2\cos 5\theta + \cos 3\theta \cos 2\theta}$ (C) $\dfrac{2\cos 5\theta - \sin 3\theta \sin 2\theta}{2\cos 5\theta + \cos 3\theta \cos 2\theta}$ (D) $\dfrac{2\cos 5\theta - \cos 3\theta \cos 2\theta}{2\cos 5\theta - \sin 3\theta \sin 2\theta}$

   ›Reveal solutionSolution

   Parametric differentiation gives $\dfrac{dy}{dx}=\dfrac{2\cos5\theta+\cos3\theta\cos2\theta}{2\cos5\theta-\sin3\theta\sin2\theta}$. Option (A).

   Solution

   With $x=\sin2\theta\cos3\theta$ and $y=\sin3\theta\cos2\theta$, differentiate each by the product rule:

   $\frac{dx}{d\theta}=2\cos2\theta\cos3\theta-3\sin2\theta\sin3\theta,\qquad \frac{dy}{d\theta}=3\cos3\theta\cos2\theta-2\sin3\theta\sin2\theta.$

   Use $\cos5\theta=\cos(3\theta+2\theta)=\cos3\theta\cos2\theta-\sin3\theta\sin2\theta$, i.e.

   $2\cos5\theta=2\cos3\theta\cos2\theta-2\sin3\theta\sin2\theta.$

   Numerator:

   $\frac{dy}{d\theta}=\bigl(2\cos3\theta\cos2\theta-2\sin3\theta\sin2\theta\bigr)+\cos3\theta\cos2\theta=2\cos5\theta+\cos3\theta\cos2\theta.$

   Denominator: …

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6. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.If $y = a \cos 3x + b e^{-x}$, then $y''(3 \sin 3x - \cos 3x) =$ (A) $10y' \sin 3x + 3y(\sin 3x + 3\cos 3x)$ (B) $10y' \cos 3x + 3y(\sin 3x + 3\cos 3x)$ (C) $10y' \cos 3x + 3y(\cos 3x + 3\sin 3x)$ (D) $10y' \cos 3x + 3y(\sin 3x - 3\cos 3x)$

   ›Reveal solutionSolution

   The key idea is to compute the first and second derivatives of $y = a \cos 3x + b e^{-x}$, then substitute into the expression $y''(3 \sin 3x - \cos 3x)$ and simplify to match one of the given forms. The result simplifies to $10y' \cos 3x + 3y(\sin 3x + 3\cos 3x)$, which corresponds to option (B).

   We start with the given function:

   $y = a \cos 3x + b e^{-x}$

   We need to find an expression for $y''(3 \sin 3x - \cos 3x)$ in terms of $y$ and $y'$. The trick is to avoid solving for $a$ and $b$ explicitly — instead, we differentiate and then cleverly combine terms.

   ---

   1. Compute the first derivative $y'$

   Differentiate term by term:

   $y' = -3a \sin 3x - b e^{-x}$

   (Recall: derivative of $\cos 3x$ is $-3\sin 3x$, and derivative of $e^{-x}$ is $-e^{-x}$.)

   2. Compute the second derivative $y''$

   Differentiate $y'$:

   $y'' = -9a \cos 3x + b e^{-x}$

   (Derivative of $-3a \sin 3x$ is $-9a \cos 3x$; derivative of $-b e^{-x}$ is $+b e^{-x}$ because $-e^{-x}$ differentiates to $e^{-x}$.)

   3. Form the expression $y''(3 \sin 3x - \cos 3x)$

   Substitute $y''$:

   $y''(3 \sin 3x - \cos 3x) = (-9a \cos 3x + b e^{-x})(3 \sin 3x - \cos 3x)$

   Expand:

   $= -27a \cos 3x \sin 3x + 9a \cos^2 3x + 3b e^{-x} \sin 3x - b e^{-x} \cos 3x$

   4. Express everything in terms of $y$ and $y'$

   We have:

   $y = a \cos 3x + b e^{-x}$

   $y' = -3a \sin 3x - b e^{-x}$

   We want to rewrite the expanded expression. Notice the terms:

   • $9a \cos^2 3x$ can be linked to $y$ and $y'$ if we also have $\sin^2$ terms, but we don't. Instead, we aim to match the pattern in the options, which involve $y' \cos 3x$ and $y(\sin 3x + 3\cos 3x)$.

   5. Try to express $a \cos 3x$ and $b e^{-x}$ from $y$ and $y'$

   From $y$ and $y'$ we can solve:

   $y = a \cos 3x + b e^{-x}$

   $y' = -3a \sin 3x - b e^{-x}$

   Add them:

   $y + y' = a \cos 3x - 3a \sin 3x = a(\cos 3x - 3\sin 3x)$

   So:

   $a = \frac{y + y'}{\cos 3x - 3\sin 3x}$

   Also subtract:

   $y - y' = a \cos 3x + 3a \sin 3x + 2b e^{-x}$

   But this gets messy. Instead, a better approach: directly compute the target expression using derivatives and compare to options.

   6. Compute $y''$ in terms of $y$ and $y'$

   Notice that from $y' = -3a \sin 3x - b e^{-x}$, we can differentiate again but also note:

   $y'' = -9a \cos 3x + b e^{-x}$

   We can write $y''$ as a combination of $y$ and $y'$:

   $y = a \cos 3x + b e^{-x}$

   $y' = -3a \sin 3x - b e^{-x}$

   Multiply $y$ by something and $y'$ by something to get $y''$. Try:

   $y'' = -9a \cos 3x + b e^{-x} = \alpha y + \beta y'$

   Substitute:

   $\alpha (a \cos 3x + b e^{-x}) + \beta (-3a \sin 3x - b e^{-x}) = a(\alpha \cos 3x - 3\beta \sin 3x) + b e^{-x}(\alpha - \beta)$

   Compare with $y'' = -9a \cos 3x + b e^{-x}$:

   • Coefficient of $a \cos 3x$: $\alpha = -9$
   • Coefficient of $a \sin 3x$: $-3\beta = 0 \Rightarrow \beta = 0$
   • Coefficient of $b e^{-x}$: $\alpha - \beta = -9 = 1$? That gives $-9 = 1$, impossible.

   So $y''$ is not a linear combination of $y$ and $y'$ alone (because the sine and cosine terms have different dependencies). So we must work directly.

   7. Instead, compute $y''(3\sin 3x - \cos 3x)$ by substituting $y''$ and then replace $a \cos 3x$ and $b e^{-x}$ using $y$ and $y'$

   We have:

   $y''(3\sin 3x - \cos 3x) = (-9a \cos 3x + b e^{-x})(3\sin 3x - \cos 3x)$

   Expand:

   $= -27a \cos 3x \sin 3x + 9a \cos^2 3x + 3b e^{-x} \sin 3x - b e^{-x} \cos 3x$

   Now, note that $y' = -3a \sin 3x - b e^{-x}$, so $3a \sin 3x = -y' - b e^{-x}$. But we have $-27a \cos 3x \sin 3x = -9 \cos 3x (3a \sin 3x) = -9 \cos 3x (-y' - b e^{-x}) = 9y' \cos 3x + 9b e^{-x} \cos 3x$.

   So the expression becomes:

   $= 9y' \cos 3x + 9b e^{-x} \cos 3x + 9a \cos^2 3x + 3b e^{-x} \sin 3x - b e^{-x} \cos 3x$

   Combine the $b e^{-x} \cos 3x$ terms: $9b e^{-x} \cos 3x - b e^{-x} \cos 3x = 8b e^{-x} \cos 3x$.

   So:

   $= 9y' \cos 3x + 9a \cos^2 3x + 3b e^{-x} \sin 3x + 8b e^{-x} \cos 3x$

   8. Now rewrite $9a \cos^2 3x$

   We know $y = a \cos 3x + b e^{-x}$, so $a \cos 3x = y - b e^{-x}$. Then:

   $9a \cos^2 3x = 9 \cos 3x (a \cos 3x) = 9 \cos 3x (y - b e^{-x}) = 9y \cos 3x - 9b e^{-x} \cos 3x$

   Substitute back:

   $= 9y' \cos 3x + (9y \cos 3x - 9b e^{-x} \cos 3x) + 3b e^{-x} \sin 3x + 8b e^{-x} \cos 3x$

   Combine the $b e^{-x} \cos 3x$: $-9 + 8 = -1$, so we get $-b e^{-x} \cos 3x$.

   Thus:

   $= 9y' \cos 3x + 9y \cos 3x + 3b e^{-x} \sin 3x - b e^{-x} \cos 3x$

   9. Factor $b e^{-x}$ terms

   Notice $3b e^{-x} \sin 3x - b e^{-x} \cos 3x = b e^{-x} (3 \sin 3x - \cos 3x)$. But $b e^{-x} = y - a \cos 3x$. However, we can also write $b e^{-x} = y - a \cos 3x$, but we want everything in $y$ and $y'$. Alternatively, note that $y' = -3a \sin 3x - b e^{-x}$ doesn't directly give $b e^{-x}$ alone.

   Instead, observe that $3y \sin 3x = 3a \cos 3x \sin 3x + 3b e^{-x} \sin 3x$. And $9y \cos 3x$ we already have. Let's try to match the options: they all have $10y' \cos 3x + 3y(\text{something})$.

   10. Compare with option (B): $10y' \cos 3x + 3y(\sin 3x + 3\cos 3x)$

   Expand option (B):

   $10y' \cos 3x + 3y \sin 3x + 9y \cos 3x$

   Our expression is:

   $9y' \cos 3x + 9y \cos 3x + 3b e^{-x} \sin 3x - b e^{-x} \cos 3x$

   We need to turn $9y' \cos 3x$ into $10y' \cos 3x$ and the remaining terms into $3y \sin 3x$. That suggests $y' \cos 3x$ appears with coefficient 10, so we might have missed a $y' \cos 3x$ term. Let's re-check step 7: we had $9y' \cos 3x$ from the manipulation, but perhaps there's an extra $y' \cos 3x$ hidden.

   11. Alternative direct substitution

   Let’s compute $y''(3\sin 3x - \cos 3x)$ by writing $y''$ in terms of $y$ and $y'$ using the original differential equation. Since $y = a \cos 3x + b e^{-x}$, note that $y$ satisfies a linear ODE. Differentiate twice: $y' = -3a \sin 3x - b e^{-x}$, $y'' = -9a \cos 3x + b e^{-x}$. Add $9y$:

   $y'' + 9y = (-9a \cos 3x + b e^{-x}) + 9(a \cos 3x + b e^{-x}) = 10 b e^{-x}$ …

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7. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.The differential equation corresponding to the family of curves $y = \log_e (ax + 3)$, where $a$ is an arbitrary constant is (A) $x \dfrac{dy}{dx} + 3e^{-x} = 1$ (B) $x \dfrac{dy}{dx} + 3e^{y} = 1$ (C) $x \dfrac{dy}{dx} + 3e^{-y} = 1$ (D) $x \dfrac{dy}{dx} + 3e^{x} = 1$

   ›Reveal solutionSolution

   The key idea is to eliminate the arbitrary constant $a$ by differentiating the given family and then substituting back. The correct differential equation is $x \frac{dy}{dx} + 3e^{-y} = 1$, which corresponds to option (C).

   We are given a family of curves $y = \log_e (ax + 3)$, where $a$ is an arbitrary constant. To find its differential equation, we need an equation involving $x$, $y$, and $\frac{dy}{dx}$ that holds for every curve in the family — meaning $a$ must be eliminated.

   The natural approach: differentiate the given relation, then use the original equation to replace $a$ in terms of $x$ and $y$.

   1. Differentiate both sides with respect to $x$. Since $y = \ln(ax + 3)$, we have

   $\frac{dy}{dx} = \frac{a}{ax + 3}.$

   2. Express $a$ from the original equation. From $y = \ln(ax + 3)$, exponentiate:

   $e^{y} = ax + 3 \quad \Rightarrow \quad ax = e^{y} - 3 \quad \Rightarrow \quad a = \frac{e^{y} - 3}{x}.$

   3. Substitute $a$ into the derivative. Replace $a$ in $\frac{dy}{dx} = \frac{a}{ax + 3}$:

   $\frac{dy}{dx} = \frac{\frac{e^{y} - 3}{x}}{e^{y}} = \frac{e^{y} - 3}{x e^{y}}.$

   4. Rearrange to match the given options. Multiply both sides by $x e^{y}$:

   $x e^{y} \frac{dy}{dx} = e^{y} - 3.$

   Bring terms together:

   $x e^{y} \frac{dy}{dx} - e^{y} = -3.$

   Factor $e^{y}$:

   $e^{y} \left( x \frac{dy}{dx} - 1 \right) = -3.$ …

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8. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.The equation of the tangent to the curve $y = \pi e^{-x/\pi}$ at the point where it crosses Y-axis is (A) $\pi x + 2y = 2\pi$ (B) $2x + \pi y = \pi^2$ (C) $x - y + \pi = 0$ (D) $x + y = \pi$

   ›Reveal solutionSolution

   To find the tangent's equation, first locate the point where the curve crosses the Y-axis, then calculate the curve's derivative at that point to get the tangent's slope. Finally, use the point-slope form. The equation of the tangent is $\boxed{x + y = \pi}$.

   The equation of a straight line, such as a tangent, can be determined if we know two things: a point it passes through and its slope. For a tangent line to a curve, the point it passes through is the point of tangency on the curve itself. The slope of the tangent at that specific point is given by the value of the derivative of the curve's equation at that point.

   Here's how we find the equation of the tangent:

   1. Identify the point of tangency:

      The problem states that the tangent is at the point where the curve $y = \pi e^{-x/\pi}$ crosses the Y-axis. A curve crosses the Y-axis when its $x$-coordinate is $0$.

      Substitute $x=0$ into the curve's equation:

      $y = \pi e^{-0/\pi}$

      $y = \pi e^0$

      Since $e^0 = 1$, we have:

      $y = \pi \times 1 = \pi$

      So, the point of tangency is $(0, \pi)$.

   2. Calculate the slope of the tangent:

      The slope of the tangent at any point $(x, y)$ on the curve is given by the derivative $\frac{dy}{dx}$.

      The curve's equation is $y = \pi e^{-x/\pi}$.

      Differentiate $y$ with respect to $x$ using the chain rule: $\frac{d}{dx}(e^{f(x)}) = e^{f(x)} \cdot f'(x)$.

      Here, $f(x) = -\frac{x}{\pi}$, so $f'(x) = -\frac{1}{\pi}$.

      $\frac{dy}{dx} = \pi \cdot \frac{d}{dx}\left(e^{-x/\pi}\right)$

      $\frac{dy}{dx} = \pi \cdot e^{-x/\pi} \cdot \left(-\frac{1}{\pi}\right)$

      $\frac{dy}{dx} = -e^{-x/\pi}$

      Now, evaluate the derivative at the point of tangency, where $x=0$:

      $m = \left.\frac{dy}{dx}\right|_{x=0} = -e^{-0/\pi} = -e^0 = -1$

      The slope of the tangent at $(0, \pi)$ is $-1$. …

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9. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
   Q.If $x = 3\sqrt{2} \cos^3 \theta$ and $y = 4 \tan^2 \theta$ then $\left(\dfrac{dy}{dx}\right)_{\theta = \pi/4} =$ (A) $\dfrac{32\sqrt{2}}{9}$ (B) $\dfrac{16}{9}$ (C) $-\dfrac{16}{9}$ (D) $-\dfrac{32}{9}$

   ›Reveal solutionSolution

   With $x=3\sqrt2\cos^3\theta,\ y=4\tan^2\theta$, the parametric derivative reduces to $\dfrac{dy}{dx}=\dfrac{-8}{9\sqrt2\,\cos^5\theta}$; at $\theta=\tfrac\pi4$ this is $-\dfrac{32}{9}$, option (D).

   1. Differentiate each parameter.

   $\frac{dx}{d\theta}=3\sqrt2\cdot3\cos^2\theta\cdot(-\sin\theta)=-9\sqrt2\cos^2\theta\sin\theta,$

   $\frac{dy}{d\theta}=4\cdot2\tan\theta\sec^2\theta=8\tan\theta\sec^2\theta.$

   2. Form the ratio.

   $\frac{dy}{dx}=\frac{8\tan\theta\sec^2\theta}{-9\sqrt2\cos^2\theta\sin\theta}.$

   Since $\tan\theta\sec^2\theta=\dfrac{\sin\theta}{\cos^3\theta}$,

   $\frac{dy}{dx}=\frac{8\,\dfrac{\sin\theta}{\cos^3\theta}}{-9\sqrt2\cos^2\theta\sin\theta}=\frac{-8}{9\sqrt2\,\cos^5\theta}.$ …

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10. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
    Q.If $x = \cos^3 \theta - \sin^3 \theta$ and $y = \sqrt[3]{\cos \theta} - \sqrt[3]{\sin \theta}$, then the value of $\dfrac{dy}{dx}$ at $\theta = \dfrac{\pi}{4}$ is (A) $\dfrac{2}{9} 3\sqrt{2}$ (B) $\dfrac{3\sqrt{2}}{3}$ (C) $\dfrac{4}{9} 3\sqrt{2}$ (D) $\dfrac{3\sqrt{2}}{9}$

    ›Reveal solutionSolution

    Differentiate $x$ and $y$ separately with respect to the parameter $\theta$ and divide. At $\theta=\pi/4$ this gives $\dfrac{dy}{dx}=\dfrac{4}{9}\cdot2^{-2/3}=\dfrac{2}{9}\sqrt[3]{2}$ — option (A).

    The concept first

    When both coordinates are given through a parameter, $x=x(\theta)$ and $y=y(\theta)$, the chain rule gives

    $\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}\qquad\left(\text{provided } \frac{dx}{d\theta}\neq0\right).$

    Never try to eliminate $\theta$ here — with a cube and a cube root in the same problem that would be brutal. Just differentiate each expression in $\theta$ and take the quotient at the required value.

    The symmetry $\sin\frac\pi4=\cos\frac\pi4=\frac{1}{\sqrt2}$ makes the arithmetic collapse very neatly, so keep the powers of $2$ in index form until the end.

    Step-by-step

    1. Differentiate $x=\cos^3\theta-\sin^3\theta$:

    $\frac{dx}{d\theta}=3\cos^2\theta(-\sin\theta)-3\sin^2\theta(\cos\theta)=-3\sin\theta\cos\theta\left(\cos\theta+\sin\theta\right).$

    2. Differentiate $y=\cos^{1/3}\theta-\sin^{1/3}\theta$:

    $\frac{dy}{d\theta}=\frac13\cos^{-2/3}\theta\,(-\sin\theta)-\frac13\sin^{-2/3}\theta(\cos\theta) =-\frac13\left(\sin\theta\,\cos^{-2/3}\theta+\cos\theta\,\sin^{-2/3}\theta\right).$

    3. Put $\theta=\dfrac{\pi}{4}$, where $\sin\theta=\cos\theta=2^{-1/2}$: $\frac{dx}{d\theta}=-3\left(\frac12\right)\left(\frac{2}{\sqrt2}\right)=-3\cdot\frac12\cdot\sqrt2=-\frac{3\sqrt2}{2}.$ …

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11. TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
    Q.The differential equation for which $y^2 = 4a(x+a)$ (where $a$ is a parameter) is general solution, is (A) $y - yy'^2 = 2xy'$ (B) $y + yy'^2 = 2xy'$ (C) $y(y + y') = 2xy'$ (D) $y(y - y') = 2xy'$

    ›Reveal solutionSolution

    To find the differential equation, we eliminate the arbitrary parameter $a$ from the given general solution $y^2 = 4a(x+a)$ by differentiating it once. The resulting differential equation is $\boxed{y - yy'^2 = 2xy'}$.

    When we are given a general solution to a differential equation, it contains one or more arbitrary constants (parameters). The process of finding the differential equation involves eliminating these arbitrary constants. The number of times we need to differentiate the given solution is equal to the number of independent arbitrary constants present in it.

    In this problem, the general solution is $y^2 = 4a(x+a)$, and $a$ is the single arbitrary parameter. Therefore, we will differentiate the equation once with respect to $x$ to obtain an expression involving $a$, and then substitute this expression back into the original equation to eliminate $a$.

    Here's the step-by-step derivation:

    1. Identify the arbitrary parameter:

       The given general solution is $y^2 = 4a(x+a)$.

       Here, $a$ is the only arbitrary parameter. Since there is one arbitrary parameter, the differential equation will be of the first order, meaning we will need to differentiate the given equation once with respect to $x$.

    2. Differentiate the equation with respect to $x$:

       We differentiate both sides of $y^2 = 4a(x+a)$ with respect to $x$. Remember to use the chain rule for $y^2$ and treat $a$ as a constant.

    $\frac{d}{dx}(y^2) = \frac{d}{dx}(4a(x+a))$

    $2y \frac{dy}{dx} = 4a \frac{d}{dx}(x+a)$

    $2y y' = 4a(1+0)$

    $2y y' = 4a$

    From this, we can express $a$ in terms of $y$ and $y'$:
    

    $a = \frac{2yy'}{4}$

    $a = \frac{yy'}{2}$

    3. Eliminate the parameter $a$: Now, substitute the expression for $a$ back into the original general solution $y^2 = 4a(x+a)$. $y^2 = 4 \left(\frac{yy'}{2}\right) \left(x + \frac{yy'}{2}\right)$ …

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12. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
    Q.If $x^2 + xy + y^2 = k$, then $\dfrac{d^2 y}{dx^2} =$ (A) $\dfrac{-6k}{(x+2y)^3}$ (B) $\dfrac{-6k}{(x+2y)^2}$ (C) $\dfrac{x^2 + xy + y^2}{(2x+y)^2}$ (D) $0$

    ›Reveal solutionSolution

    Implicit differentiation twice gives $\dfrac{d^2y}{dx^2}=\dfrac{-6k}{(x+2y)^3}$ — option (A).

    Step 1 — First derivative. Differentiate $x^2+xy+y^2=k$:

    $2x+y+x y'+2y y'=0 \;\Rightarrow\; y'=-\frac{2x+y}{x+2y}.$

    Step 2 — Second derivative. With $y'=-\dfrac{u}{v}$ where $u=2x+y,\ v=x+2y$ (so $u'=2+y',\ v'=1+2y'$):

    $y''=-\frac{u'v-uv'}{v^2}.$

    Compute the numerator:

    $u'v-uv'=(2+y')(x+2y)-(2x+y)(1+2y')=3y-3x\,y'.$

    Substitute $y'=-\dfrac{2x+y}{x+2y}$: …

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