Find $\dfrac{dy}{dx}$ if $x=\cos^{-1}\dfrac{2t}{1+t^2}$, $y=\sec^{-1}\left(\sqrt{1+t^2}\right)$.

Find (dy)/(dx) if x=cos⁻¹(2t)/(1+t²), y=sec⁻¹(√(1+t²)).

EXERCISE 1.4 · Q157

\dfrac{dy}{dx}

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★

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We have $x=\cos^{-1}\dfrac{2t}{1+t^2}$ , $y=\sec^{-1}\sqrt{1+t^2}$ . Put $t=\tan\theta$ .

Step 1. Since $\dfrac{2t}{1+t^2}=\sin2\theta$ ,

x=\cos^{-1}(\sin2\theta)=\cos^{-1}\!\left[\cos\left(\frac{\pi}{2}-2\theta\right)\right]=\frac{\pi}{2}-2\theta

(for $\theta$ in the range where $\frac{\pi}{2}-2\theta\in[0,\pi]$ ), so $\dfrac{dx}{d\theta}=-2$ , and with $\theta=\tan^{-1}t$ , $\dfrac{d\theta}{dt}=\dfrac{1}{1+t^2}$ , giving $\dfrac{dx}{dt}=\dfrac{-2}{1+t^2}$ . …

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Q.Find $\dfrac{dy}{dx}$ if $x=\cos^{-1}\dfrac{2t}{1+t^2}$ , $y=\sec^{-1}\left(\sqrt{1+t^2}\right)$ .

Concept understanding — Derivatives of Parametric Functions

Q.Find dydx\dfrac{dy}{dx}dxdy​ if x=cos⁡−12t1+t2x=\cos^{-1}\dfrac{2t}{1+t^2}x=cos−11+t22t​, y=sec⁡−1(1+t2)y=\sec^{-1}\left(\sqrt{1+t^2}\right)y=sec−1(1+t2​).

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Q.Find $\dfrac{dy}{dx}$ if $x=\cos^{-1}\dfrac{2t}{1+t^2}$ , $y=\sec^{-1}\left(\sqrt{1+t^2}\right)$ .

Unlock everything free for 14 days