Question

If \(x = \sqrt{a^{sin^{-1} t}}\), \(y = \sqrt{a^{ cos^{-1}t}}\) then \(\frac{dy}{dx}\) is:

• A. \(\frac{y}{x}\)
• B. \(-\frac{y}{x}\)
• C. \(-\frac{y^2}{x}\)
• D. \(-\frac{y}{x^2}\)

Answer 1

The Correct Option is B

To find \(\frac{dy}{dx}\) given \(x = \sqrt{a^{\sin^{-1} t}}\) and \(y = \sqrt{a^{\cos^{-1} t}}\), we start by first expressing \(x\) and \(y\) in terms of \(t\). Since \(x = \sqrt{a^{\sin^{-1} t}}\), we have:

\(x = a^{\frac{1}{2}\sin^{-1} t}\)

Similarly, \(y = \sqrt{a^{\cos^{-1} t}}\) becomes:

\(y = a^{\frac{1}{2}\cos^{-1} t}\)

To find \(\frac{dy}{dx}\), we use the chain rule by first finding \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

For \(\ln x\):

\(\ln x = \frac{1}{2} \sin^{-1} t \cdot \ln a\)

Differentiate with respect to \(t\):

\(\frac{1}{x}\cdot \frac{dx}{dt} = \frac{1}{2 \sqrt{1 - t^2}} \cdot \ln a\)

Thus,

\(\frac{dx}{dt} = x \cdot \frac{\ln a}{2 \sqrt{1 - t^2}}\)

For \(\ln y\):

\(\ln y = \frac{1}{2} \cos^{-1} t \cdot \ln a\)

Differentiate with respect to \(t\):

\(\frac{1}{y} \cdot \frac{dy}{dt} = -\frac{1}{2 \sqrt{1 - t^2}} \cdot \ln a\)

Thus,

\(\frac{dy}{dt} = y \cdot -\frac{\ln a}{2 \sqrt{1 - t^2}}\)

Using the chain rule, we have:

\(\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}\)

Substitute \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\):

\(\frac{dy}{dx} = \frac{y \cdot -\frac{\ln a}{2 \sqrt{1 - t^2}}}{x \cdot \frac{\ln a}{2 \sqrt{1 - t^2}}}\)

This simplifies to:

\(\frac{dy}{dx} = -\frac{y}{x}\)

Therefore, the correct answer is \(-\frac{y}{x}\).