Q.Differentiate w.r.t. $x$: $\tan^{-1}\left(\dfrac{3a^2 x - x^3}{a^3 - 3ax^2}\right),\ -\dfrac{1}{\sqrt{3}} < \dfrac{x}{a} < \dfrac{1}{\sqrt{3}}$.

Showing the 12 most recent of 34 on this concept.

1. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
   Q.If $y = \mathrm{Tan}^{-1}\left\{\dfrac{ax-b}{bx+a}\right\}$, then $y' =$ _______ (A) $\dfrac{1}{1+x^2} + \dfrac{a^2}{a^2+b^2}$ (B) $\dfrac{1}{1+x^2}$ (C) $\dfrac{1}{1+\left(\dfrac{ax-b}{bx+a}\right)^2}$ (D) $\dfrac{bx+a}{1+(ax-b)^2}$

   ›Reveal solutionSolution

   This tests differentiating an arctangent of a Möbius-type expression, which (as often happens) collapses to the simple form $\tfrac{1}{1+x^2}$. Answer: $\dfrac{1}{1+x^2}$.

   Concept and Intuition

   Expressions of the form $\mathrm{Tan}^{-1}\!\left(\dfrac{ax-b}{bx+a}\right)$ often equal $\mathrm{Tan}^{-1}(x) - \mathrm{Tan}^{-1}(b/a)$ up to a constant (since $\dfrac{ax-b}{bx+a} = \dfrac{\frac{a}{b}x - 1}{x+\frac ab}$ resembles $\tan(\alpha-\beta)$ with a constant angle $\beta$), so its derivative should reduce to just $\dfrac{1}{1+x^2}$ — the constant angle contributes zero derivative.

   Step-by-Step Solution

   1. Let $u = \dfrac{ax-b}{bx+a}$. By the quotient rule, $u' = \dfrac{a(bx+a) - (ax-b)b}{(bx+a)^2} = \dfrac{abx+a^2-abx+b^2}{(bx+a)^2} = \dfrac{a^2+b^2}{(bx+a)^2}$.
   2. Compute $1+u^2 = 1+\dfrac{(ax-b)^2}{(bx+a)^2} = \dfrac{(bx+a)^2+(ax-b)^2}{(bx+a)^2}$. …

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2. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.If $y = \tan^{-1}\left(\dfrac{a\cos x - b\sin x}{b\cos x + a\sin x}\right)$, then $\dfrac{dy}{dx} =$ ______ (A) $0$ (B) $\dfrac{a}{b}$ (C) $-1$ (D) $2$

   ›Reveal solutionSolution

   Tests recognizing an $a\cos x - b\sin x$ / $b\cos x + a\sin x$ ratio as a shifted cotangent using the auxiliary-angle (R-method) substitution, collapsing the arctan to a linear function of $x$.

   Concept and Intuition

   Expressions like $a\cos x - b\sin x$ can always be written as $r\cos(x+\phi)$ for suitable $r=\sqrt{a^2+b^2}$ and angle $\phi$ with $a=r\cos\phi$, $b=r\sin\phi$. Once both numerator and denominator are expressed this way, their ratio collapses to a simple trig ratio in $(x+\phi)$, and the $\arctan$ of that becomes an explicit linear expression in $x$ — making differentiation trivial.

   Step-by-Step Solution

   1. Let $a=r\cos\phi$, $b=r\sin\phi$ where $r=\sqrt{a^2+b^2}$.
   2. Numerator: $a\cos x - b\sin x = r\cos\phi\cos x - r\sin\phi\sin x = r\cos(x+\phi)$.
   3. Denominator: $b\cos x + a\sin x = r\sin\phi\cos x + r\cos\phi\sin x = r\sin(x+\phi)$.
   4. Ratio: $\dfrac{r\cos(x+\phi)}{r\sin(x+\phi)}=\cot(x+\phi)=\tan\left(\dfrac\pi2-(x+\phi)\right)$. …

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3. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
   Q.$\dfrac{d}{dx}\left(\operatorname{Tan}^{-1}\left(\dfrac{\cos x}{1+\sin x}\right)\right)=$ (A) $\dfrac{1}{2}$ (B) $\dfrac{-1}{2}$ (C) $1$ (D) $-1$

   ›Reveal solutionSolution

   Simplify the argument to a single tangent of a half-angle expression, then differentiate. Answer: $-\dfrac12$.

   Concept and Intuition

   Expressions like $\dfrac{\cos x}{1+\sin x}$ are classic half-angle simplifications that collapse to $\tan\left(\dfrac\pi4-\dfrac x2\right)$, letting the inverse tangent cancel with the tangent directly.

   Step-by-Step Solution

   1. Multiply numerator and denominator by $(1-\sin x)$: $\dfrac{\cos x}{1+\sin x}=\dfrac{\cos x(1-\sin x)}{1-\sin^2x}=\dfrac{\cos x(1-\sin x)}{\cos^2x}=\dfrac{1-\sin x}{\cos x}$.
   2. This is a known identity: $\dfrac{1-\sin x}{\cos x}=\tan\left(\dfrac\pi4-\dfrac x2\right)$. …

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4. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If $y = \mathrm{Tan}^{-1}\left(\dfrac{x}{1+2x^2}\right) + \mathrm{Tan}^{-1}\left(\dfrac{x}{1+6x^2}\right)$, then $\dfrac{dy}{dx}=$ (A) $\dfrac{4}{16x^2+1} - \dfrac{3}{9x^2+1}$ (B) $\dfrac{3}{9x^2+1} - \dfrac{1}{x^2+1}$ (C) $\dfrac{3}{9x^2+1} - \dfrac{2}{4x^2+1}$ (D) $\dfrac{1}{9x^2+1} - \dfrac{1}{x^2+1}$

   ›Reveal solutionSolution

   Recognizing each arctan term as a telescoping difference $\tan^{-1}(2x)-\tan^{-1}(x)$ and $\tan^{-1}(3x)-\tan^{-1}(2x)$ collapses $y$ to $\tan^{-1}3x-\tan^{-1}x$, whose derivative is immediate.

   Concept and Intuition

   Terms of the form $\tan^{-1}\!\left(\dfrac{a-b}{1+ab}\right)$ are exactly $\tan^{-1}a - \tan^{-1}b$ (the tangent subtraction identity, valid when $ab>-1$). Spotting this pattern turns an awkward-looking sum into a telescoping simplification, avoiding messy direct differentiation of nested rational-argument arctans.

   Step-by-Step Solution

   1. Compare $\dfrac{x}{1+2x^2}$ to the form $\dfrac{a-b}{1+ab}$: try $a=2x,\,b=x$, giving $\dfrac{2x-x}{1+2x\cdot x}=\dfrac{x}{1+2x^2}$ ✓. So

   $\tan^{-1}\!\left(\frac{x}{1+2x^2}\right) = \tan^{-1}(2x)-\tan^{-1}(x)$

   2. Compare $\dfrac{x}{1+6x^2}$ similarly: try $a=3x,\,b=2x$, giving $\dfrac{3x-2x}{1+3x\cdot2x}=\dfrac{x}{1+6x^2}$ ✓. So

   $\tan^{-1}\!\left(\frac{x}{1+6x^2}\right) = \tan^{-1}(3x)-\tan^{-1}(2x)$

   3. Add the two:

   $y = \big[\tan^{-1}2x-\tan^{-1}x\big] + \big[\tan^{-1}3x-\tan^{-1}2x\big] = \tan^{-1}3x - \tan^{-1}x$

   (the $\tan^{-1}2x$ terms cancel — a telescoping sum).

   4. Differentiate: …

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5. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.If $y = \tan^{-1}\left(\dfrac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}}\right)$, where $x^2 \le 1$. Then find $\dfrac{dy}{dx}$ (A) $\dfrac{\pi}{4} + \dfrac{1}{2}\cos^{-1}(x^2)$ (B) $\dfrac{\pi}{4} - \dfrac{1}{2}\cos^{-1}(x^2)$ (C) $\dfrac{-x}{\sqrt{1-x^4}}$ (D) $\dfrac{-2x}{\sqrt{1-x^4}}$

   ›Reveal solutionSolution

   Substituting $x^2=\cos\varphi$ collapses the expression to $y=\pi/4+\tfrac12\cos^{-1}(x^2)$, whose derivative is $\dfrac{-x}{\sqrt{1-x^4}}$.

   Concept and Intuition

   Nested-radical inverse-trig expressions like this are almost always designed to simplify via a trig substitution that turns $\sqrt{1\pm x^2}$ into $\sqrt2\cos$ or $\sqrt2\sin$ of a half-angle, converting the whole ratio into a single tangent — much easier to differentiate than the raw radical expression.

   Step-by-Step Solution

   1. Let $x^2=\cos\varphi$ (valid since $x^2\le1$). Then $1+x^2=2\cos^2(\varphi/2)$ and $1-x^2=2\sin^2(\varphi/2)$.
   2. So $\sqrt{1+x^2}=\sqrt2\cos(\varphi/2)$ and $\sqrt{1-x^2}=\sqrt2\sin(\varphi/2)$.
   3. The ratio becomes $\dfrac{\cos(\varphi/2)+\sin(\varphi/2)}{\cos(\varphi/2)-\sin(\varphi/2)} = \dfrac{1+\tan(\varphi/2)}{1-\tan(\varphi/2)} = \tan\!\left(\dfrac{\pi}{4}+\dfrac{\varphi}{2}\right)$.
   4. So $y=\tan^{-1}\!\left[\tan\!\left(\dfrac\pi4+\dfrac\varphi2\right)\right]=\dfrac\pi4+\dfrac\varphi2=\dfrac\pi4+\dfrac12\cos^{-1}(x^2)$. …

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6. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If $\dfrac{1}{2}\sin^{-1}\left(\dfrac{3\sin 2\theta}{5+4\cos 2\theta}\right) = \tan^{-1}x$ then $x =$ (A) $\tan\dfrac{\theta}{3}$ (B) $\dfrac{1}{3}\tan\theta$ (C) $\tan 3\theta$ (D) $\dfrac{1}{3}\tan 3\theta$

   ›Reveal solutionSolution

   Rewrite the inverse-sine argument as $\sin(2\phi)$ for a suitable $\tan\phi$ built from $\tan\theta$, so the half of the inverse sine collapses to $\tan^{-1}$ of that quantity directly. Answer: $x=\dfrac13\tan\theta$.

   Concept and Intuition

   Expressions of the form $\dfrac{k\sin2\theta}{A+B\cos2\theta}$ can often be massaged, after dividing through by $(1+\tan^2\theta)$ i.e. converting to $t=\tan\theta$, into the recognizable double-angle sine pattern $\dfrac{2u}{1+u^2}=\sin(2\tan^{-1}u)$ for some rescaled variable $u$. Spotting this pattern converts an awkward inverse-trig expression into a clean angle.

   Step-by-Step Solution

   1. Let $t=\tan\theta$. Recall $\sin2\theta=\dfrac{2t}{1+t^2}$, $\cos2\theta=\dfrac{1-t^2}{1+t^2}$.
   2. Numerator: $3\sin2\theta=\dfrac{6t}{1+t^2}$.
   3. Denominator: $5+4\cos2\theta=5+\dfrac{4(1-t^2)}{1+t^2}=\dfrac{5(1+t^2)+4(1-t^2)}{1+t^2}=\dfrac{9+t^2}{1+t^2}$.
   4. So the ratio is $\dfrac{6t/(1+t^2)}{(9+t^2)/(1+t^2)}=\dfrac{6t}{9+t^2}$.
   5. Let $u=t/3$. Then $\dfrac{2u}{1+u^2}=\dfrac{2t/3}{1+t^2/9}=\dfrac{2t/3}{(9+t^2)/9}=\dfrac{6t}{9+t^2}$ — exactly matches!
   6. So $\dfrac{6t}{9+t^2}=\sin(2\phi)$ where $\tan\phi=u=t/3$, i.e. $\phi=\tan^{-1}\!\left(\dfrac{\tan\theta}{3}\right)$. …

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7. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.If $y=\text{Tan}^{-1}\sqrt{\dfrac{1-\cos x}{1+\cos x}}$, then the values of $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ respectively are ________ (A) $1,\ 0$ (B) $\dfrac{x}{2},\ \dfrac{1}{2}$ (C) $\dfrac{1}{2},\ 0$ (D) $\dfrac{-1}{2},\ 0$

   ›Reveal solutionSolution

   The half-angle identity collapses $y$ to $x/2$, so $dy/dx = 1/2$ and $d^2y/dx^2 = 0$.

   Concept and Intuition

   Expressions like $\sqrt{\dfrac{1-\cos x}{1+\cos x}}$ are classic half-angle simplifications; recognizing the identity turns an intimidating inverse-trig derivative problem into a trivial linear function.

   Step-by-Step Solution

   1. Recall $1-\cos x = 2\sin^2(x/2)$ and $1+\cos x = 2\cos^2(x/2)$.
   2. So $\dfrac{1-\cos x}{1+\cos x} = \tan^2(x/2)$, and $\sqrt{\tan^2(x/2)} = |\tan(x/2)|$.
   3. On the principal branch where $\tan(x/2)\ge 0$ (i.e. $x\in(-\pi,\pi)$), $y = \operatorname{Tan}^{-1}(\tan(x/2)) = \dfrac{x}{2}$.
   4. Differentiate: $\dfrac{dy}{dx} = \dfrac12$.
   5. Differentiate again: since $dy/dx$ is constant, $\dfrac{d^2y}{dx^2} = 0$.

   Common Mistakes …

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8. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.If $y=\text{Tan}^{-1}\dfrac{x}{1+2x^2}+\text{Tan}^{-1}\dfrac{x}{1+6x^2}+\text{Tan}^{-1}\dfrac{x}{1+12x^2}$, then $\left(\dfrac{dy}{dx}\right)_{x=\frac{1}{2}}=$ (A) $1$ (B) $-1$ (C) $0$ (D) $\frac{1}{2}$

   ›Reveal solutionSolution

   Recognize the telescoping arctan pattern to collapse $y$ to $\arctan(4x)-\arctan(x)$; the derivative at $x=\frac12$ is $0$.

   Concept and Intuition

   The identity $\arctan A-\arctan B=\arctan\dfrac{A-B}{1+AB}$ (mod branch issues) means a sum like $\arctan\dfrac{x}{1+n(n+1)x^2}$, recognized as $\arctan((n+1)x)-\arctan(nx)$, telescopes when summed over consecutive $n$ — a huge simplification before ever differentiating.

   Step-by-Step Solution

   1. Check the general term: $\arctan((n+1)x)-\arctan(nx)=\arctan\dfrac{(n+1)x-nx}{1+n(n+1)x^2}=\arctan\dfrac{x}{1+n(n+1)x^2}$.
   2. Match given terms: $\dfrac{x}{1+2x^2}$ has $n(n+1)=2\Rightarrow n=1$; $\dfrac{x}{1+6x^2}$ has $n(n+1)=6\Rightarrow n=2$; $\dfrac{x}{1+12x^2}$ has $n(n+1)=12\Rightarrow n=3$.
   3. So $y=[\arctan2x-\arctan x]+[\arctan3x-\arctan2x]+[\arctan4x-\arctan3x]=\arctan4x-\arctan x$ (everything else cancels). …

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9. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
   Q.If $y = \log_{\cot x} \tan x - \log_{\tan x} \cot x + \tan^{-1}\left(\dfrac{4x}{4-x^2}\right)$, then $\dfrac{dy}{dx} =$ ______ (A) $\dfrac{1}{4+x^2}$ (B) $\dfrac{4}{4+x^2}$ (C) $\dfrac{1}{4-x^2}$ (D) $\dfrac{4}{4-x^2}$

   ›Reveal solutionSolution

   This tests the change-of-base log identity and the double-angle form of $\tan^{-1}$. The two log terms cancel completely, and the answer is $\dfrac{4}{4+x^2}$.

   Concept and Intuition

   Whenever you see $\log_a b$ and $\log_b a$ together, remember they are reciprocals of each other: $\log_a b=1/\log_b a$. Here $a=\cot x,\,b=\tan x$ are reciprocals of each other too, so $\ln(\tan x)=-\ln(\cot x)$, which forces both log terms to equal $-1$ and cancel. What's left is the classic $\tan^{-1}\left(\dfrac{2t}{1-t^2}\right)=2\tan^{-1}t$ substitution pattern with $t=x/2$.

   Step-by-Step Solution

   1. $\log_{\cot x}\tan x=\dfrac{\ln\tan x}{\ln\cot x}=\dfrac{\ln\tan x}{\ln(1/\tan x)}=\dfrac{\ln\tan x}{-\ln\tan x}=-1$.
   2. Similarly $\log_{\tan x}\cot x=-1$.
   3. So $y=-1-(-1)+\tan^{-1}\!\left(\dfrac{4x}{4-x^2}\right)=\tan^{-1}\!\left(\dfrac{4x}{4-x^2}\right)$. …

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10. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
    Q.For $0 < x < 1$, $\displaystyle\int [\mathrm{Tan}^{-1}(1 - x + x^2) + \mathrm{Tan}^{-1}(1 - x)]\, dx =$ (A) $x\,\mathrm{Cot}^{-1}x + \log\sqrt{1+x^2} + c$ (B) $x\,\mathrm{Tan}^{-1}x - \log(1+x^2) + c$ (C) $x\,\mathrm{Cot}^{-1}x + \dfrac{3}{4}\log(1+x^2) + c$ (D) $x\,\mathrm{Tan}^{-1}x - \dfrac{3}{4}\log\sqrt{1+x^2} + c$

    ›Reveal solutionSolution

    The two arctangent terms combine, via the tan-addition identity, into a single $\mathrm{Cot}^{-1}x$; integrating that by parts gives option (A).

    Concept and Intuition

    The stem looks intimidating because it has two separate inverse-tangent terms with messy arguments. The key insight is that $\mathrm{Tan}^{-1}p+\mathrm{Tan}^{-1}q$ always collapses via

    $\mathrm{Tan}^{-1}p+\mathrm{Tan}^{-1}q=\mathrm{Tan}^{-1}\!\left(\frac{p+q}{1-pq}\right)\ (\text{mod }\pi\text{ correction}),$

    so it's worth testing whether $p=1-x+x^2$ and $q=1-x$ are designed to make $\dfrac{p+q}{1-pq}$ simplify beautifully — which they are.

    Step-by-Step Solution

    1. Compute $p+q = (1-x+x^2)+(1-x) = 2-2x+x^2$.
    2. Compute $pq = (1-x+x^2)(1-x)$. Expanding: $(1-x+x^2)(1-x) = 1-2x+2x^2-x^3$.
    3. So $1-pq = 1-(1-2x+2x^2-x^3) = 2x-2x^2+x^3 = x(2-2x+x^2)$.
    4. Hence $\dfrac{p+q}{1-pq} = \dfrac{2-2x+x^2}{x(2-2x+x^2)} = \dfrac{1}{x}$.
    5. Check the correction term: for $0<x<1$, $2-2x+x^2=(x-1)^2+1>0$ and $x>0$, so $1-pq>0 \Rightarrow pq<1$, meaning the plain addition formula applies with no $\pm\pi$ shift.
    6. So the integrand is exactly $\mathrm{Tan}^{-1}(1/x) = \mathrm{Cot}^{-1}x$ (valid since $x>0$).
    7. Now integrate by parts: $\displaystyle\int \mathrm{Cot}^{-1}x\,dx = x\,\mathrm{Cot}^{-1}x - \int x\cdot\left(\frac{-1}{1+x^2}\right)dx = x\,\mathrm{Cot}^{-1}x+\int\frac{x}{1+x^2}dx$.
    8. $\displaystyle\int\frac{x}{1+x^2}dx = \frac12\log(1+x^2)=\log\sqrt{1+x^2}$.
    9. Total: $x\,\mathrm{Cot}^{-1}x+\log\sqrt{1+x^2}+c$. …

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11. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
    Q.$\tan\left(2\,\mathrm{Tan}^{-1}\left(\dfrac{1}{3}\right) + \mathrm{Tan}^{-1}\left(\dfrac{1}{7}\right)\right) =$ (A) $\dfrac{1}{\sqrt{3}}$ (B) $\sqrt{3}$ (C) $1$ (D) $3/7$

    ›Reveal solutionSolution

    First reduce the double-angle inverse-tangent term to a single tangent value using the tangent double-angle formula, then combine with the second $\tan^{-1}$ term using the tangent addition formula. The result is exactly $1$.

    Concept and Intuition

    Expressions like $\tan(2\tan^{-1}x+\tan^{-1}y)$ are handled in two stages: first collapse $2\tan^{-1}x$ to a single angle whose tangent is known via the double-angle formula $\tan2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha}$, then treat the whole thing as $\tan(\alpha'+\beta)$ using the standard addition formula, where $\alpha'$ is the angle with $\tan\alpha'=\tan(2\tan^{-1}x)$.

    Step-by-Step Solution

    1. Let $\alpha=\tan^{-1}(1/3)$, so $\tan\alpha=1/3$. Then $\tan2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha}=\dfrac{2/3}{1-1/9}=\dfrac{2/3}{8/9}=\dfrac{2}{3}\times\dfrac{9}{8}=\dfrac{18}{24}=\dfrac34$.
    2. Let $\beta=\tan^{-1}(1/7)$, so $\tan\beta=1/7$.
    3. We need $\tan(2\alpha+\beta)=\dfrac{\tan2\alpha+\tan\beta}{1-\tan2\alpha\tan\beta}=\dfrac{\frac34+\frac17}{1-\frac34\cdot\frac17}$. …

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12. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
    Q.$\mathrm{Tanh}^{-1}\left(\dfrac{1}{3}\right) + \mathrm{Coth}^{-1}(3) =$ (A) $\mathrm{Sech}^{-1}\left(\dfrac{1}{3}\right)$ (B) $\mathrm{Cosech}^{-1}\left(\dfrac{1}{3}\right)$ (C) $\mathrm{Cosh}^{-1}\left(\dfrac{4}{3}\right)$ (D) $\mathrm{Sinh}^{-1}\left(\dfrac{3}{4}\right)$

    ›Reveal solutionSolution

    This tests the identity linking $\mathrm{Coth}^{-1}$ to $\mathrm{Tanh}^{-1}$ and the logarithmic form of inverse hyperbolic functions; the sum collapses to $\log 2 = \mathrm{Sinh}^{-1}(3/4)$.

    Concept and Intuition

    Inverse hyperbolic functions all reduce to logarithms. For $|x|>1$, $\mathrm{Coth}^{-1}(x) = \mathrm{Tanh}^{-1}(1/x)$ because $\coth\theta = x \iff \tanh\theta = 1/x$. This lets us rewrite both terms of the sum using the SAME inverse function, so they simply add.

    Step-by-Step Solution

    1. Since $3>1$, use $\mathrm{Coth}^{-1}(3) = \mathrm{Tanh}^{-1}(1/3)$.
    2. The sum becomes $\mathrm{Tanh}^{-1}(1/3) + \mathrm{Tanh}^{-1}(1/3) = 2\,\mathrm{Tanh}^{-1}(1/3)$.
    3. Use $\mathrm{Tanh}^{-1}(y) = \dfrac12 \log\!\left(\dfrac{1+y}{1-y}\right)$ with $y = 1/3$: $\mathrm{Tanh}^{-1}(1/3) = \dfrac12\log\!\left(\dfrac{4/3}{2/3}\right) = \dfrac12 \log 2$.
    4. So the sum $= 2\times \dfrac12\ln2 = \log 2$.
    5. Test $\mathrm{Sinh}^{-1}(3/4) = \log\!\left(y+\sqrt{y^2+1}\right)$ with $y=3/4$: $\log\!\left(\dfrac34+\sqrt{\dfrac{9}{16}+1}\right) = \log\!\left(\dfrac34+\dfrac54\right) = \log 2$. Exact match. …

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