Question

find dy for ( y = e^{sqrt{x}} - 2 ).
for ( y = e^{sqrt{x}} - 2 ), ( dy = ) dx.
(type an exact answer, using radicals as needed.)

Explanation:

Step1: Differentiate the function using the chain rule

Let \( u = \sqrt{x}-2=x^{\frac{1}{2}} - 2\). Then \( y = e^{u}\).
The derivative of \(y\) with respect to \(u\) is \(\frac{dy}{du}=e^{u}\), and the derivative of \(u\) with respect to \(x\) is \(\frac{du}{dx}=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}\).
By the chain rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\).
Substituting \(u = \sqrt{x}-2\) into \(\frac{dy}{du}\), we get \(\frac{dy}{du}=e^{\sqrt{x}-2}\).
So \(\frac{dy}{dx}=e^{\sqrt{x}-2}\cdot\frac{1}{2\sqrt{x}}\).

Step2: Find \(dy\)

Since \(dy=\frac{dy}{dx}dx\), substituting \(\frac{dy}{dx}=e^{\sqrt{x}-2}\cdot\frac{1}{2\sqrt{x}}\) into \(dy=\frac{dy}{dx}dx\), we have \(dy = \frac{e^{\sqrt{x}-2}}{2\sqrt{x}}dx\).

Answer:

\(\frac{e^{\sqrt{x}-2}}{2\sqrt{x}}\)