Differentiation by substitution Questions in English

(C) The general term of the series is $T_n = \tan^{-1}\left(\frac{1}{\sin^2 x + (2n-1)\sin x + (n^2-n+1)}\right)$.
We can rewrite this as $T_n = \tan^{-1}\left(\frac{(\sin x + n) - (\sin x + n - 1)}{1 + (\sin x + n)(\sin x + n - 1)}\right)$.
Using the identity $\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$,we get $T_n = \tan^{-1}(\sin x + n) - \tan^{-1}(\sin x + n - 1)$.
Summing up to $10$ terms,we have $f(x) = \sum_{n=1}^{10} [\tan^{-1}(\sin x + n) - \tan^{-1}(\sin x + n - 1)]$.
This is a telescoping series,so $f(x) = \tan^{-1}(\sin x + 10) - \tan^{-1}(\sin x)$.
Differentiating with respect to $x$,we get $f'(x) = \frac{\cos x}{1 + (\sin x + 10)^2} - \frac{\cos x}{1 + \sin^2 x}$.
At $x = 0$,$f'(0) = \frac{\cos 0}{1 + (0 + 10)^2} - \frac{\cos 0}{1 + 0^2} = \frac{1}{101} - 1 = \frac{-100}{101}$.

(C) Let $y = \tan ^{-1}\left[\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}\right]$.
We know that $1 + \sin x = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$.
Similarly,$1 - \sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$.
Substituting these into the expression:
$y = \tan ^{-1}\left[\frac{(\cos \frac{x}{2} + \sin \frac{x}{2}) - (\cos \frac{x}{2} - \sin \frac{x}{2})}{(\cos \frac{x}{2} + \sin \frac{x}{2}) + (\cos \frac{x}{2} - \sin \frac{x}{2})}\right]$
$y = \tan ^{-1}\left[\frac{2 \sin \frac{x}{2}}{2 \cos \frac{x}{2}}\right] = \tan ^{-1}(\tan \frac{x}{2}) = \frac{x}{2}$.
Now,differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\frac{x}{2}) = \frac{1}{2}$.

(D) Given $y=\cos ^{-1}\left(\frac{6 x-2 x^2-4}{2 x^2-6 x+5}\right)$.
Let $v = \frac{6 x-2 x^2-4}{2 x^2-6 x+5}$.
Note that $v = \frac{-(2x^2-6x+5)+1}{2x^2-6x+5} = -1 + \frac{1}{2x^2-6x+5}$.
Using the chain rule,$\frac{dy}{dx} = \frac{-1}{\sqrt{1-v^2}} \cdot \frac{dv}{dx}$.
Calculating $\frac{dv}{dx}$ using the quotient rule:
$\frac{dv}{dx} = \frac{(6-4x)(2x^2-6x+5) - (6x-2x^2-4)(4x-6)}{(2x^2-6x+5)^2} = \frac{2(3-2x)(2x^2-6x+5) + 2(2x^2-6x+4)(2x-3)}{(2x^2-6x+5)^2} = \frac{2(2x-3)[-(2x^2-6x+5) + (2x^2-6x+4)]}{(2x^2-6x+5)^2} = \frac{2(2x-3)(-1)}{(2x^2-6x+5)^2} = \frac{-2(2x-3)}{(2x^2-6x+5)^2}$.
Now,$1-v^2 = 1 - \left(\frac{6x-2x^2-4}{2x^2-6x+5}\right)^2 = \frac{(2x^2-6x+5)^2 - (6x-2x^2-4)^2}{(2x^2-6x+5)^2} = \frac{(2x^2-6x+5 - 6x+2x^2+4)(2x^2-6x+5 + 6x-2x^2-4)}{(2x^2-6x+5)^2} = \frac{(4x^2-12x+9)(1)}{(2x^2-6x+5)^2} = \frac{(2x-3)^2}{(2x^2-6x+5)^2}$.
Thus,$\sqrt{1-v^2} = \frac{|2x-3|}{2x^2-6x+5}$.
Substituting back: $\frac{dy}{dx} = \frac{-1}{\frac{|2x-3|}{2x^2-6x+5}} \cdot \frac{-2(2x-3)}{(2x^2-6x+5)^2} = \frac{2(2x-3)}{|2x-3|(2x^2-6x+5)}$.
Assuming $2x-3 > 0$,$\frac{dy}{dx} = \frac{2}{2x^2-6x+5}$.

(D) Given $y=\tan ^{-1}\left[\frac{\sin ^3(2 x)-3 x^2 \sin (2 x)}{3 x \sin ^2(2 x)-x^3}\right]$.
Divide the numerator and denominator by $x^3$:
$y=\tan ^{-1}\left[\frac{(\frac{\sin 2x}{x})^3 - 3(\frac{\sin 2x}{x})}{3(\frac{\sin 2x}{x})^2 - 1}\right]$.
Let $\frac{\sin 2x}{x} = \tan \theta$. Then $y = \tan^{-1} \left[ \frac{\tan^3 \theta - 3 \tan \theta}{3 \tan^2 \theta - 1} \right]$.
Using the identity $\tan 3\theta = \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}$,we have $\frac{\tan^3 \theta - 3 \tan \theta}{3 \tan^2 \theta - 1} = -\tan 3\theta$.
So,$y = \tan^{-1}(-\tan 3\theta) = -3\theta = -3 \tan^{-1}(\frac{\sin 2x}{x})$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = -3 \cdot \frac{1}{1 + (\frac{\sin 2x}{x})^2} \cdot \frac{d}{dx}(\frac{\sin 2x}{x})$.
$\frac{d}{dx}(\frac{\sin 2x}{x}) = \frac{x(2 \cos 2x) - \sin 2x}{x^2}$.
$\frac{dy}{dx} = -3 \cdot \frac{x^2}{x^2 + \sin^2 2x} \cdot \frac{2x \cos 2x - \sin 2x}{x^2} = \frac{3 \sin 2x - 6x \cos 2x}{x^2 + \sin^2 2x}$.
Note: The expression inside the bracket is $-\tan 3\theta$,so $y = -3\theta$. The derivative is $\frac{3 \sin 2x - 6x \cos 2x}{x^2 + \sin^2 2x}$. Comparing with options,option $D$ is the intended form if the sign inside the bracket was inverted.

(B) Given,$y = \tan^{-1} \left[ \frac{5 \cos x - 12 \sin x}{12 \cos x + 5 \sin x} \right]$.
Divide the numerator and denominator by $12 \cos x$:
$y = \tan^{-1} \left[ \frac{\frac{5}{12} - \tan x}{1 + \frac{5}{12} \tan x} \right]$.
Using the formula $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1} \left( \frac{A - B}{1 + AB} \right)$,we get:
$y = \tan^{-1} \left( \frac{5}{12} \right) - \tan^{-1}(\tan x)$.
$y = \tan^{-1} \left( \frac{5}{12} \right) - x$.
Differentiating both sides with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} \left( \tan^{-1} \left( \frac{5}{12} \right) \right) - \frac{d}{dx}(x)$.
Since $\tan^{-1} \left( \frac{5}{12} \right)$ is a constant,its derivative is $0$.
$\frac{dy}{dx} = 0 - 1 = -1$.

(A) Given: $y = \tan^{-1} \left( \frac{\sqrt{1+x^2}-1}{x} \right)$.
Substitute $x = \tan \theta$,which implies $\theta = \tan^{-1} x$.
Then,$y = \tan^{-1} \left( \frac{\sqrt{1+\tan^2 \theta}-1}{\tan \theta} \right) = \tan^{-1} \left( \frac{\sec \theta - 1}{\tan \theta} \right)$.
Using trigonometric identities,$\frac{\sec \theta - 1}{\tan \theta} = \frac{1-\cos \theta}{\sin \theta} = \frac{2 \sin^2 (\theta/2)}{2 \sin(\theta/2) \cos(\theta/2)} = \tan(\theta/2)$.
So,$y = \tan^{-1} (\tan(\theta/2)) = \theta/2 = \frac{1}{2} \tan^{-1} x$.
Differentiating with respect to $x$,we get $y' = \frac{1}{2(1+x^2)}$.
At $x = 1$,$y'(1) = \frac{1}{2(1+1^2)} = \frac{1}{2(2)} = \frac{1}{4}$.

(A) Given $y = \tan^{-1} \sqrt{\frac{1-\sin x}{1+\sin x}}$.
Using the identity $\sin x = \cos(\frac{\pi}{2}-x)$,we have:
$y = \tan^{-1} \sqrt{\frac{1-\cos(\frac{\pi}{2}-x)}{1+\cos(\frac{\pi}{2}-x)}}$.
Using half-angle formulas $1-\cos \theta = 2\sin^2(\frac{\theta}{2})$ and $1+\cos \theta = 2\cos^2(\frac{\theta}{2})$:
$y = \tan^{-1} \sqrt{\frac{2\sin^2(\frac{\pi}{4}-\frac{x}{2})}{2\cos^2(\frac{\pi}{4}-\frac{x}{2})}} = \tan^{-1} \sqrt{\tan^2(\frac{\pi}{4}-\frac{x}{2})}$.
Assuming the principal value range,$y = \frac{\pi}{4} - \frac{x}{2}$.
Now,differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{4} - \frac{x}{2}) = 0 - \frac{1}{2} = -\frac{1}{2}$.
Thus,the value at $x = \frac{\pi}{6}$ is $-\frac{1}{2}$.

(B) Let $y = \tan^{-1} \left( \frac{\cos x}{1 + \sin x} \right)$.
Using the trigonometric identities $\cos x = \sin(\frac{\pi}{2} - x)$ and $1 + \sin x = 1 + \cos(\frac{\pi}{2} - x) = 2 \cos^2(\frac{\pi}{4} - \frac{x}{2})$:
$\frac{\cos x}{1 + \sin x} = \frac{\sin(\frac{\pi}{2} - x)}{2 \cos^2(\frac{\pi}{4} - \frac{x}{2})} = \frac{2 \sin(\frac{\pi}{4} - \frac{x}{2}) \cos(\frac{\pi}{4} - \frac{x}{2})}{2 \cos^2(\frac{\pi}{4} - \frac{x}{2})} = \tan(\frac{\pi}{4} - \frac{x}{2})$.
Thus,$y = \tan^{-1} \left( \tan(\frac{\pi}{4} - \frac{x}{2}) \right) = \frac{\pi}{4} - \frac{x}{2}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{4} - \frac{x}{2} \right) = -\frac{1}{2}$.

(C) Let $y = y_1 + y_2$,where $y_1 = \tan^{-1} \left( \frac{3\cos x - 4\sin x}{4\cos x + 3\sin x} \right)$ and $y_2 = 2\tan^{-1} \left( \frac{x}{1+\sqrt{1-x^2}} \right)$.
For $y_1$,divide numerator and denominator by $4\cos x$: $y_1 = \tan^{-1} \left( \frac{3/4 - \tan x}{1 + (3/4)\tan x} \right) = \tan^{-1}(3/4) - x$.
Thus,$\frac{dy_1}{dx} = -1$.
For $y_2$,let $x = \sin \theta$,then $\theta = \sin^{-1} x$. The expression becomes $2\tan^{-1} \left( \frac{\sin \theta}{1+\cos \theta} \right) = 2\tan^{-1} \left( \tan(\theta/2) \right) = \theta = \sin^{-1} x$.
Thus,$\frac{dy_2}{dx} = \frac{1}{\sqrt{1-x^2}}$.
Therefore,$\frac{dy}{dx} = -1 + \frac{1}{\sqrt{1-x^2}}$.
At $x = \frac{\sqrt{3}}{2}$,$\frac{dy}{dx} = -1 + \frac{1}{\sqrt{1 - 3/4}} = -1 + \frac{1}{\sqrt{1/4}} = -1 + 2 = 1$.