Question

1. calculator allowed if \\( \frac{dy}{dx} = \sqrt{1 - y^2} \\), then \\( \frac{d^2y}{dx^2} = \\)

Explanation:

Step1: Recall the chain rule for differentiation

To find the second derivative \(\frac{d^2y}{dx^2}\), we need to differentiate the first derivative \(\frac{dy}{dx}\) with respect to \(x\). Given \(\frac{dy}{dx}=\sqrt{1 - y^{2}}=(1 - y^{2})^{\frac{1}{2}}\), we will use the chain rule. The chain rule states that if we have a function \(u(v(x))\), then the derivative with respect to \(x\) is \(u^\prime(v(x))\cdot v^\prime(x)\). Here, let \(u = v^{\frac{1}{2}}\) where \(v=1 - y^{2}\), and we also need to remember that \(y\) is a function of \(x\), so when we differentiate \(y\) with respect to \(x\), we have \(\frac{dy}{dx}\) (which we already know).

Step2: Differentiate \(\frac{dy}{dx}\) with respect to \(x\)

First, find the derivative of \(u = v^{\frac{1}{2}}\) with respect to \(v\): \(\frac{du}{dv}=\frac{1}{2}v^{-\frac{1}{2}}=\frac{1}{2\sqrt{v}}=\frac{1}{2\sqrt{1 - y^{2}}}\) (since \(v = 1 - y^{2}\)).

Then, find the derivative of \(v = 1 - y^{2}\) with respect to \(x\). Using the chain rule, \(\frac{dv}{dx}=\frac{d}{dx}(1)-\frac{d}{dx}(y^{2})=0 - 2y\frac{dy}{dx}=- 2y\frac{dy}{dx}\).

Now, by the chain rule, \(\frac{d^2y}{dx^2}=\frac{du}{dv}\cdot\frac{dv}{dx}\). Substitute the values we found:

\(\frac{d^2y}{dx^2}=\frac{1}{2\sqrt{1 - y^{2}}}\cdot(-2y\frac{dy}{dx})\)

Simplify the expression: The \(2\) in the numerator and denominator cancels out, so we get \(\frac{d^2y}{dx^2}=\frac{-y\frac{dy}{dx}}{\sqrt{1 - y^{2}}}\)

Step3: Substitute \(\frac{dy}{dx}=\sqrt{1 - y^{2}}\) into the expression

We know that \(\frac{dy}{dx}=\sqrt{1 - y^{2}}\), so substitute this into the above formula:

\(\frac{d^2y}{dx^2}=\frac{-y\cdot\sqrt{1 - y^{2}}}{\sqrt{1 - y^{2}}}\)

The \(\sqrt{1 - y^{2}}\) in the numerator and denominator cancels out (assuming \(\sqrt{1 - y^{2}}
eq0\), if \(\sqrt{1 - y^{2}} = 0\), we can check separately, but in general, for the domain where the original derivative is defined, we can do this cancellation). So we are left with \(\frac{d^2y}{dx^2}=-y\)

Answer:

\(-y\)