Question

if (f(x)=e^{(2/x)}), then (f(x)=)
(a) (2e^{(2/x)}ln x)
(b) (e^{(2/x)})
(c) (e^{(- 2/x^{2})})
(d) (-\frac{2}{x^{2}}e^{(2/x)})
(e) (-2x^{2}e^{(2/x)})

Explanation:

Step1: Identify the outer - inner functions

Let $u = \frac{2}{x}=2x^{-1}$, then $y = e^{u}$. The derivative of $y = e^{u}$ with respect to $u$ is $\frac{dy}{du}=e^{u}$, and the derivative of $u = 2x^{-1}$ with respect to $x$ is $\frac{du}{dx}=- 2x^{-2}=-\frac{2}{x^{2}}$.

Step2: Apply the chain - rule

The chain - rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substituting $\frac{dy}{du}=e^{u}$ and $\frac{du}{dx}=-\frac{2}{x^{2}}$ into the chain - rule formula, and since $u = \frac{2}{x}$, we get $\frac{dy}{dx}=e^{\frac{2}{x}}\cdot(-\frac{2}{x^{2}})=-\frac{2}{x^{2}}e^{\frac{2}{x}}$.

Answer:

D. $-\frac{2}{x^{2}}e^{\frac{2}{x}}$