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

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.)

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.)

Answer

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}})