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Question

Mathematics Question on Differentiability

If x=2cosec−1tx = \sqrt{2^{cosec^{-1} } t} and y=2sec⁡−1t(∣t∣≥1)y = \sqrt{2^{\sec^{-1}} t} (|t | \geq 1), then dydx\frac{dy}{dx} is equal to :

A

yx\frac{y}{x}

B

xy\frac{x}{y}

C

−yx - \frac{y}{x}

D

−xy - \frac{x}{y}

Answer

−yx - \frac{y}{x}

Explanation

Solution

Given: x=2cose −1tx=\sqrt{2^{\text {cose }^{-1} t}} and y=2sec −1t(∣t∣≥1)y=\sqrt{2^{\text {sec }^{-1} t}}(|t| \geq 1)
Now, dydx=dy/dtdx/dt\frac{d y}{d x}=\frac{d y / d t}{d x / d t}
=122sec⁡−1t2sec⁡−1tln⁡(1tt2−1)−122cosec−1t2cosec −1ln⁡(1tt2−1)=\frac{\frac{1}{2 \sqrt{2 \sec ^{-1} t}} 2^{\sec ^{-1} t} \ln \left(\frac{1}{t \sqrt{t^{2}-1}}\right)}{-\frac{1}{2 \sqrt{2 cosec^{-1} t}} 2^{\text {cosec }^{-1}} \ln \left(\frac{1}{t \sqrt{t^{2}-1}}\right)}
=−2ec−1t2omee −1y=−yx=-\frac{\sqrt{2^{ ec ^{-1} t}}}{\sqrt{2^{\text {omee }^{-1} y}}}=\frac{-y}{x}