Question

Find the differential \(d y\) of the given function. $$ y=\sqrt{x}+1 / \sqrt{x} $$

Short answer

The differential of the function \(y = \sqrt{x} + \frac{1}{\sqrt{x}}\) is \(dy = \left(\frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}}\right)dx\).

Step-by-step solution

Rewrite the function

Begin by rewriting the function in a form that will be easier to differentiate. The expression \(\sqrt{x}\) is the same as \(x^{1/2}\) and \(1 / \sqrt{x} = x^{-1/2}\). So, the function is rewritten as \(y = x^{1/2} + x^{-1/2}\).

Find the derivative of the function

Next, use power rule to differentiate the function. The derivative of \(x^n\) with respect to \(x\) is \(n x^{n-1}\). \Hence, the derivative \(y'\) of the function is \(y' = \frac{1}{2} x^{1/2 - 1} - \frac{1}{2} x^{-1/2 - 1} = \frac{1}{2} x^{-1/2} - \frac{1}{2} x^{-3/2} = \frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}}\)

Find the differential

The differential of the function, \(dy\), is given by the derivative of the function times \(dx\). So, \(dy = y' dx = \left(\frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}}\right)dx\).