Question

If \(y=sin^{-1}x \) and \((1-x^2)\frac{d^2y}{dx^2} -x \frac{dy}{dx}=K\),then value of K is:

• A. 2
• B. \(\frac12\)
• C. 0
• D. 1

Answer 1

The Correct Option is C

Given that \(y=\sin^{-1}x\), we need to find \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\).

We know the derivative of \(\sin^{-1}x\) is \(\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}\).

Now, let's differentiate \(\frac{dy}{dx}\) with respect to \(x\) to find \(\frac{d^2y}{dx^2}\):

\(\frac{d}{dx}\left(\frac{1}{\sqrt{1-x^2}}\right) = \frac{d}{dx}(1-x^2)^{-\frac{1}{2}}\).

Using the chain rule, this becomes:

\(-\frac{1}{2}(1-x^2)^{-\frac{3}{2}}(0-2x) = \frac{x}{(1-x^2)^{\frac{3}{2}}}\).

Hence, \(\frac{d^2y}{dx^2}=\frac{x}{(1-x^2)^{\frac{3}{2}}}\).

We substitute these into the given equation:

\((1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx}=K\).

\((1-x^2)\left(\frac{x}{(1-x^2)^{\frac{3}{2}}}\right) - x\left(\frac{1}{\sqrt{1-x^2}}\right)=K\).

Simplifying this, we get:

\(\frac{x}{\sqrt{1-x^2}}-\frac{x}{\sqrt{1-x^2}}=K\).

This simplifies to \(0=K\).

Thus, the value of \(K\) is 0.