Question

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

Explanation:

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

Let \(u=\sqrt{x}=x^{\frac{1}{2}}\). Then \(y = e^{u}-2\).
By the chain - rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\).
First, find \(\frac{dy}{du}\):
Since \(y = e^{u}-2\), then \(\frac{dy}{du}=e^{u}\).
Second, find \(\frac{du}{dx}\):
Since \(u = x^{\frac{1}{2}}\), then \(\frac{du}{dx}=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}\).

Step2: Calculate \(\frac{dy}{dx}\)

By the chain - rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=e^{u}\cdot\frac{1}{2\sqrt{x}}\).
Substitute \(u = \sqrt{x}\) back in, we get \(\frac{dy}{dx}=\frac{e^{\sqrt{x}}}{2\sqrt{x}}\).
Since \(dy=\frac{dy}{dx}dx\).

Answer:

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