Question

If \(xsiny + ysinx = \pi\), then \(\frac{dy}{dx}\) at \(\)\((\frac{\pi}{2},\frac{\pi}{2})\)is equal to

• A. 1
• B. \(\frac{\pi}{2}\)
• C. -1
• D. \(\frac{-\pi}{2}\)
• E. 0

Answer 1

The Correct Option is C

We are given the equation: \[ x \sin y + y \sin x = \pi. \] We need to find the value of \( \frac{d y}{d x} \) at \( \left( \frac{\pi}{2}, \frac{\pi}{2} \right) \).
Step 1: Differentiate implicitly with respect to \( x \) Differentiate both sides of the equation \( x \sin y + y \sin x = \pi \) with respect to \( x \).
The derivative of \( x \sin y \) using the product rule: \[ \frac{d}{dx}(x \sin y) = \sin y + x \cos y \frac{dy}{dx}. \]
The derivative of \( y \sin x \) using the product rule: \[ \frac{d}{dx}(y \sin x) = \sin x \frac{dy}{dx} + y \cos x. \]
The derivative of \( \pi \) is 0 since it is a constant.
Thus, the differentiated equation is: \[ \sin y + x \cos y \frac{dy}{dx} + \sin x \frac{dy}{dx} + y \cos x = 0. \]
Step 2: Solve for \( \frac{dy}{dx} \) Rearrange the equation to isolate \( \frac{dy}{dx} \): \[ x \cos y \frac{dy}{dx} + \sin x \frac{dy}{dx} = -(\sin y + y \cos x). \] Factor out \( \frac{dy}{dx} \): \[ \frac{dy}{dx} \left( x \cos y + \sin x \right) = -(\sin y + y \cos x). \] Solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{-(\sin y + y \cos x)}{x \cos y + \sin x}. \]
Step 3: Substitute \( \left( \frac{\pi}{2}, \frac{\pi}{2} \right) \) into the equation Substitute \( x = \frac{\pi}{2} \) and \( y = \frac{\pi}{2} \) into the equation.
First, calculate the terms:
\( \sin \left( \frac{\pi}{2} \right) = 1 \),
\( \cos \left( \frac{\pi}{2} \right) = 0 \),
\( \sin \left( \frac{\pi}{2} \right) = 1 \),
\( \cos \left( \frac{\pi}{2} \right) = 0 \).
Substituting these values into the equation: \[ \frac{dy}{dx} = \frac{-(1 + \frac{\pi}{2} \cdot 0)}{\frac{\pi}{2} \cdot 0 + 1} = \frac{-1}{1} = -1. \]

So, the correct option is (C) : \(-1\)

Answer 2

• We are given: \[x\sin y + y\sin x = \pi\]
• Differentiate both sides with respect to \(x\), using the product rule: \[\frac{d}{dx}(x\sin y) + \frac{d}{dx}(y\sin x) = \frac{d}{dx}(\pi)\]
• Apply the product rule: \[(\sin y + x\cos y \frac{dy}{dx}) + (\sin x \frac{dy}{dx} + y\cos x) = 0\]
• Rearrange to solve for \(\frac{dy}{dx}\): \[x\cos y \frac{dy}{dx} + \sin x \frac{dy}{dx} = -\sin y - y\cos x\] \[\frac{dy}{dx} (x\cos y + \sin x) = -(\sin y + y\cos x)\] \[\frac{dy}{dx} = \frac{-(\sin y + y\cos x)}{x\cos y + \sin x}\]
• Now, we want to evaluate \(\frac{dy}{dx}\) at \((\frac{\pi}{2}, \frac{\pi}{2})\). Substitute \(x = \frac{\pi}{2}\) and \(y = \frac{\pi}{2}\) into the expression: \[\frac{dy}{dx} = \frac{-(\sin(\frac{\pi}{2}) + \frac{\pi}{2}\cos(\frac{\pi}{2}))}{\frac{\pi}{2}\cos(\frac{\pi}{2}) + \sin(\frac{\pi}{2})}\]
• Evaluate the trigonometric functions: \[\sin(\frac{\pi}{2}) = 1\] \[\cos(\frac{\pi}{2}) = 0\]
• Substitute these values: \[\frac{dy}{dx} = \frac{-(1 + \frac{\pi}{2} \cdot 0)}{\frac{\pi}{2} \cdot 0 + 1} = \frac{-1}{1} = -1\]
• Therefore, \(\frac{dy}{dx}\) at \((\frac{\pi}{2}, \frac{\pi}{2})\) is equal to -1.