How do you find the derivative of y=e^(1/x)?

Oct 15, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {e}^{\frac{1}{x}} / {x}^{2}$

Explanation:

Use the chain rule by posing $u = \frac{1}{x}$, so that $y = {e}^{u}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}} = \frac{d}{\mathrm{du}} {e}^{u} \times \frac{d}{\mathrm{dx}} \frac{1}{x} = {e}^{u} \times - \frac{1}{x} ^ 2 = - {e}^{\frac{1}{x}} / {x}^{2}$

Oct 15, 2017

Let's see.

Explanation:

Let the equation $y = {e}^{\frac{1}{x}}$ be a function of $x$ such that $\rightarrow$

$y = f \left(x\right) = {e}^{\frac{1}{x}}$

Now, differentiating the equation w.r.t $x$ we get $\rightarrow$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({e}^{\frac{1}{x}}\right)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{\frac{1}{x}} \cdot \frac{d}{\mathrm{dx}} \left(\frac{1}{x}\right)$ $\leftarrow$ Chain Rule.

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{\frac{1}{x}} \cdot \left(- \frac{1}{x} ^ 2\right)$

$\therefore \textcolor{red}{\frac{\mathrm{dy}}{\mathrm{dx}} = - {e}^{\frac{1}{x}} / {x}^{2}}$. (Answer).

Hope it Helps:)