Question

If \(x=\sqrt{10^{\cos^{-1}\theta}}\ and\ y=\sqrt{10^{\sin^{-1}\theta}}\), then \(\frac{dy}{dx}\) is equal to

• A. xy
• B. \(\frac{x}{y}\)
• C. \(\frac{y}{x}\)
• D. \(\frac{-x}{y}\)
• E. \(\frac{-y}{x}\)

Answer 1

Answer: (E)

We are given the functions: \[ x = \sqrt{10 \cos^{-1} \theta}, \quad y = \sqrt{10 \sin^{-1} \theta}. \] To find \( \frac{dy}{dx} \), we will use the chain rule. First, differentiate \( x \) and \( y \) with respect to \( \theta \). For \( x = \sqrt{10 \cos^{-1} \theta} \), we differentiate with respect to \( \theta \): \[ \frac{dx}{d\theta} = \frac{1}{2} \cdot 10^{-1/2} \cdot \frac{d}{d\theta} (\cos^{-1} \theta) = \frac{1}{2} \cdot \frac{10^{1/2}}{1 - \theta^2}. \] Similarly, for \( y = \sqrt{10 \sin^{-1} \theta} \), we differentiate: \[ \frac{dy}{d\theta} = \frac{1}{2} \cdot 10^{-1/2} \cdot \frac{d}{d\theta} (\sin^{-1} \theta) = \frac{1}{2} \cdot \frac{10^{1/2}}{\sqrt{1 - \theta^2}}. \] Now, \( \frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta} \): \[ \frac{dy}{dx} = \frac{\frac{1}{2} \cdot \frac{10^{1/2}}{\sqrt{1 - \theta^2}}}{\frac{1}{2} \cdot \frac{10^{1/2}}{1 - \theta^2}} = \frac{-y}{x}. \]

So, the correct option is (E) : \(\frac{-y}{x}\)

Answer 2

• We are given: \[x=\sqrt{10^{\cos^{-1}\theta}} = 10^{\frac{1}{2}\cos^{-1}\theta}\] \[y=\sqrt{10^{\sin^{-1}\theta}} = 10^{\frac{1}{2}\sin^{-1}\theta}\]
• Take the natural logarithm of both equations: \[\ln x = \frac{1}{2} \cos^{-1}\theta \cdot \ln 10\] \[\ln y = \frac{1}{2} \sin^{-1}\theta \cdot \ln 10\]
• Differentiate both equations with respect to \(\theta\): \[\frac{1}{x}\frac{dx}{d\theta} = \frac{1}{2} \ln 10 \cdot \frac{-1}{\sqrt{1-\theta^2}}\] \[\frac{1}{y}\frac{dy}{d\theta} = \frac{1}{2} \ln 10 \cdot \frac{1}{\sqrt{1-\theta^2}}\]
• Solve for \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\): \[\frac{dx}{d\theta} = \frac{-x\ln 10}{2\sqrt{1-\theta^2}}\] \[\frac{dy}{d\theta} = \frac{y\ln 10}{2\sqrt{1-\theta^2}}\]
• Now, we want to find \(\frac{dy}{dx}\), which is \(\frac{dy/d\theta}{dx/d\theta}\): \[\frac{dy}{dx} = \frac{\frac{y\ln 10}{2\sqrt{1-\theta^2}}}{\frac{-x\ln 10}{2\sqrt{1-\theta^2}}} = \frac{y\ln 10}{2\sqrt{1-\theta^2}} \cdot \frac{2\sqrt{1-\theta^2}}{-x\ln 10}\]
• Simplify the expression: \[\frac{dy}{dx} = \frac{y}{-x} = -\frac{y}{x}\]
• Therefore, \(\frac{dy}{dx} = -\frac{y}{x}\).