Question

if y = x - x^(-1), then the value of dy/dx at the point (0, 1) is a - 1 b 1/2 c 1 + 2x d undefined

Explanation:

Step1: Differentiate \(y = x^{2}-x^{-1}\)

Using the power - rule \(\frac{d}{dx}(x^{n})=nx^{n - 1}\), we have \(\frac{dy}{dx}=2x+x^{-2}=2x+\frac{1}{x^{2}}\).

Step2: Evaluate at \(x = 1\)

Substitute \(x = 1\) into \(\frac{dy}{dx}\). When \(x = 1\), \(\frac{dy}{dx}=2\times1+\frac{1}{1^{2}}=2 + 1=3\).

Answer:

None of the given options are correct. The value of \(\frac{dy}{dx}\) at the point \((1,0)\) (since when \(x = 1\), \(y=1^{2}-1^{-1}=0\)) for \(y=x^{2}-x^{-1}\) is \(3\).