Differentiate: $y=\sin^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)$

Differentiate: y=sin⁻¹((1-x²)/(1+x²))

Question 117 of 143

y=\sin^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)

Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2019Subjective· 3mImportance★★★★★

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Concept understanding — Derivatives of Standard Functions

Applying the first-principle limit once to each basic elementary function builds a permanent table; every later problem then differentiates by combining this table with the rules of differentiation (sum/product/quotient/chain/constant-multiple), with no further limit ever required.

Algebraic functions.

\frac{d}{dx}(k)=0\ (k\text{ constant}), \qquad \frac{d}{dx}\!\left(x^n\right) = nx^{n-1}\ \text{for any real }n\ \text{(Corollaries 10.1–10.2 extend the integer case to rational, then any real, exponent).}

Logarithmic and exponential functions.

\frac{d}{dx}(\log x) = \frac1x,\qquad \frac{d}{dx}(\log_a x)=\frac{1}{x\log a},\qquad \frac{d}{dx}\!\left(a^x\right) = a^x\log a,\qquad \frac{d}{dx}\!\left(e^x\right)=e^x.

( $e^x$ is the unique elementary function that is its own derivative — the special case $a=e$ , $\log e=1$ .)

The six trigonometric functions.

\frac{d}{dx}(\sin x)=\cos x,\qquad \frac{d}{dx}(\cos x)=-\sin x,\qquad \frac{d}{dx}(\tan x)=\sec^2x,

\frac{d}{dx}(\sec x)=\sec x\tan x,\qquad \frac{d}{dx}(\operatorname{cosec}x)=-\operatorname{cosec}x\cot x,\qquad \frac{d}{dx}(\cot x)=-\operatorname{cosec}^2x.

Only $\sin x\to\cos x$ is derived directly from the limit definition (via the sum-to-product identity and $\lim_{\theta\to0}\sin\theta/\theta=1$ ); every other trig derivative follows from it using the chain rule ( $\cos x=\sin(\tfrac\pi2+x)$ ) or the quotient rule ( $\tan x=\sin x/\cos x$ , etc.) — so the whole trig table rests on a single limit.

The six inverse trigonometric functions (each domain-restricted to its principal branch):

$\frac{d}{dx}\!\left(\sin^{-1}x\right)=\frac{1}{\sqrt{1-x^2}},\qquad \frac{d}{dx}\!\left(\cos^{-1}x\right)=-\frac{1}{\sqrt{1-x^2}},\qquad \frac{d}{dx}\!\left(\tan^{-1}x\right)=\frac{1}{1+x^2},$ …

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With $x=\tan\theta$ , the expression becomes $\sin^{-1}(\cos2\theta)=\dfrac{\pi}{2}-2\theta$ (for $x>0$ ), which differentiates directly to $-\dfrac{2}{1+x^2}$ .

Let $x=\tan\theta$ , so $\theta=\tan^{-1}x$ .

$\dfrac{1-x^2}{1+x^2} = \dfrac{1-\tan^2\theta}{1+\tan^2\theta} = \cos2\theta$ (standard identity).

So $y = \sin^{-1}(\cos2\theta)$ . For $x>0$ (so $\theta\in(0,\pi/2)$ , giving $\pi/2-2\theta\in(-\pi/2,\pi/2)$ , within the principal range of $\sin^{-1}$ ), $\cos2\theta = \sin\left(\dfrac{\pi}{2}-2\theta\right)$ , so $y=\dfrac{\pi}{2}-2\theta = \dfrac{\pi}{2}-2\tan^{-1}x$ .

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Q.Differentiate: y=sin⁡−1(1−x21+x2)y=\sin^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)y=sin−1(1+x21−x2​)

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Q.Differentiate: $y=\sin^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)$

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