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

Question

Let $f(x)=|\log_e x|-|x-1|+5$. Statement 1: $f(x)$ is differentiable for all $x\in(0,\infty)$
Statement 2: $f(x)$ is increasing in $(1,\infty)$
Statement 3: $f(x)$ is decreasing in $(0,1)$
Which of the following is correct?

Answer 1

To solve the problem, we need to find $\frac{dy}{dx}$ given $x = \sqrt{2^{\csc^{-1} t}}$ and $y = \sqrt{2^{\sec^{-1} t}}$ where $|t| \geq 1$.

Start with the expression for $x$: $$x = \sqrt{2^{\csc^{-1} t}}$$. By expressing $\csc^{-1}$ in terms of $t$, we have: $$\csc^{-1} t = \theta \Rightarrow \csc \theta = t \Rightarrow \sin \theta = \frac{1}{t}$$. Hence, $x = (2^{\theta/2})$.
Now, for $y$: $$y = \sqrt{2^{\sec^{-1} t}}$$. Expressing $\sec^{-1}$ in terms of $t$, we have: $$\sec^{-1} t = \phi \Rightarrow \sec \phi = t \Rightarrow \cos \phi = \frac{1}{t}$$. Hence, $y = (2^{\phi/2})$.
Notice that: $$\theta = \csc^{-1} t \text{ and } \phi = \sec^{-1} t$$ are inverse trigonometric functions related to angle representations. Observing the derivatives of these: $$\frac{d}{dt}(\csc^{-1} t) = -\frac{1}{|t|\sqrt{t^2 - 1}}$$ and $$\frac{d}{dt}(\sec^{-1} t) = \frac{1}{|t|\sqrt{t^2 - 1}}$$.
The expressions for $x$ and $y$ become: $$ x = 2^{\theta/2} $$ and $$ y = 2^{\phi/2} $$.
Given the relation: $$\theta = \csc^{-1} t$$ and $$\phi = \sec^{-1} t$$, differentiate both to find $\frac{dx}{dt}$ and $\frac{dy}{dt}$.
From the derivatives found in earlier steps, $$\frac{dx}{dt} = \frac{-\ln 2 \times 2^{\theta/2}}{2|t|\sqrt{t^2 - 1}}$$ and $$\frac{dy}{dt} = \frac{\ln 2 \times 2^{\phi/2}}{2|t|\sqrt{t^2 - 1}}$$.
Since $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$, substitute the derivatives: $$\frac{dy}{dx} = \frac{\frac{\ln 2 \times 2^{\phi/2}}{2|t|\sqrt{t^2 - 1}}}{\frac{-\ln 2 \times 2^{\theta/2}}{2|t|\sqrt{t^2 - 1}}}$$. Simplifying, we get: $$\frac{dy}{dx} = - \frac{2^{\phi/2}}{2^{\theta/2}} = - \frac{y}{x}$$.
Therefore, the value of $\frac{dy}{dx}$ is $ - \frac{y}{x}$, confirming the correct answer choice as specified.

Hence, the final answer is $ - \frac{y}{x}$.

Answer 2

To solve the problem, we need to find $\frac{dy}{dx}$ given $x = \sqrt{2^{\csc^{-1} t}}$ and $y = \sqrt{2^{\sec^{-1} t}}$ where $|t| \geq 1$.

Start with the expression for $x$: $$x = \sqrt{2^{\csc^{-1} t}}$$. By expressing $\csc^{-1}$ in terms of $t$, we have: $$\csc^{-1} t = \theta \Rightarrow \csc \theta = t \Rightarrow \sin \theta = \frac{1}{t}$$. Hence, $x = (2^{\theta/2})$.
Now, for $y$: $$y = \sqrt{2^{\sec^{-1} t}}$$. Expressing $\sec^{-1}$ in terms of $t$, we have: $$\sec^{-1} t = \phi \Rightarrow \sec \phi = t \Rightarrow \cos \phi = \frac{1}{t}$$. Hence, $y = (2^{\phi/2})$.
Notice that: $$\theta = \csc^{-1} t \text{ and } \phi = \sec^{-1} t$$ are inverse trigonometric functions related to angle representations. Observing the derivatives of these: $$\frac{d}{dt}(\csc^{-1} t) = -\frac{1}{|t|\sqrt{t^2 - 1}}$$ and $$\frac{d}{dt}(\sec^{-1} t) = \frac{1}{|t|\sqrt{t^2 - 1}}$$.
The expressions for $x$ and $y$ become: $$ x = 2^{\theta/2} $$ and $$ y = 2^{\phi/2} $$.
Given the relation: $$\theta = \csc^{-1} t$$ and $$\phi = \sec^{-1} t$$, differentiate both to find $\frac{dx}{dt}$ and $\frac{dy}{dt}$.
From the derivatives found in earlier steps, $$\frac{dx}{dt} = \frac{-\ln 2 \times 2^{\theta/2}}{2|t|\sqrt{t^2 - 1}}$$ and $$\frac{dy}{dt} = \frac{\ln 2 \times 2^{\phi/2}}{2|t|\sqrt{t^2 - 1}}$$.
Since $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$, substitute the derivatives: $$\frac{dy}{dx} = \frac{\frac{\ln 2 \times 2^{\phi/2}}{2|t|\sqrt{t^2 - 1}}}{\frac{-\ln 2 \times 2^{\theta/2}}{2|t|\sqrt{t^2 - 1}}}$$. Simplifying, we get: $$\frac{dy}{dx} = - \frac{2^{\phi/2}}{2^{\theta/2}} = - \frac{y}{x}$$.
Therefore, the value of $\frac{dy}{dx}$ is $ - \frac{y}{x}$, confirming the correct answer choice as specified.

Hence, the final answer is $ - \frac{y}{x}$.

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

The Correct Option is C

Solution and Explanation

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