# How do you differentiate f(x)=tan(1/x)  using the chain rule?

Apr 5, 2018

$f ' \left(x\right) = - {\sec}^{2} \frac{\frac{1}{x}}{x} ^ 2$

#### Explanation:

$f \left(x\right) = \tan \left(\frac{1}{x}\right)$

$f \left(x\right) = \tan \left({x}^{-} 1\right)$

Now, when applying the Chain Rule to the tangent function, where $u$ is a function,

$\frac{d}{\mathrm{dx}} \tan \left(u\right) = {\sec}^{2} \left(u\right) \frac{\mathrm{du}}{\mathrm{dx}}$

Thus,

$f ' \left(x\right) = {\sec}^{2} \left(\frac{1}{x}\right) \cdot \frac{d}{\mathrm{dx}} {x}^{-} 1$

$f ' \left(x\right) = - {\sec}^{2} \frac{\frac{1}{x}}{x} ^ 2$