Question

first derivative and second derivative of f(x)= e^(1/x)

Answer 1

first derivative and second derivative of $f(x) = e^{\wedge} (1 / x)$

The product rule:

$$(f * g) ^ {\prime} = g * f ^ {\prime} + f * g ^ {\prime}$$

The derivative of the function of a function $h(x) = f(g(x))$ with respect to $x$ is:

$$h (x) ^ {\prime} = f ^ {\prime} (g (x)) * g ^ {\prime} (x)$$

derivative of exponent:

$$\frac {d}{d x} (e ^ {x}) = e ^ {x}$$

derivative of $x^n$ :

$$\frac {d}{d x} (x ^ {n}) = n x ^ {n - 1}$$

So, the first derivative equals:

$$f ^ {\prime} (x) = \frac {d}{d x} \left(e ^ {\frac {1}{x}}\right) = e ^ {\frac {1}{x}} \frac {d}{d x} \left(\frac {1}{x}\right) = e ^ {\frac {1}{x}} \left(- \frac {1}{x ^ {2}}\right) = - \frac {e ^ {\frac {1}{x}}}{x ^ {2}}$$

and the second derivative:

$$f ^ {\prime \prime} (x) = \frac {d}{d x} \left(- \frac {e ^ {\frac {1}{x}}}{x ^ {2}}\right) = - \left(\frac {1}{x ^ {2}} \frac {d}{d x} \left(e ^ {\frac {1}{x}}\right) + e ^ {\frac {1}{x}} \frac {d}{d x} \left(\frac {1}{x ^ {2}}\right)\right) = - \left(- \frac {e ^ {\frac {1}{x}}}{x ^ {4}} - \frac {2 e ^ {\frac {1}{x}}}{x ^ {3}}\right) = \frac {e ^ {\frac {1}{x}}}{x ^ {4}} (1 + 2 x)$$

Answer: the first derivative $= -\frac{e^{\frac{1}{x}}}{x^{2}}$ , the second derivative $= \frac{e^{\frac{1}{x}}}{x^{4}} (1 + 2x)$