How do you differentiate f(x)=tan(lnx)  using the chain rule?

Feb 16, 2016

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

Explanation:

The chain rule states that

d/dx(f(g(x))=f'(g(x))*g'(x)

First, we must know that the derivative of $\tan \left(x\right)$ is ${\sec}^{2} \left(x\right)$. With this knowledge, we can create a version of the chain rule specific to tangent functions:

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

=>d/dx(tan(g(x))=sec^2(g(x))*g'(x)

So, if we are differentiating the function $f \left(x\right) = \tan \left(\ln \left(x\right)\right)$, we see that

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

We must now know that the derivative of $\ln \left(x\right)$ is $1 \text{/} x$, yielding the simplified derivative:

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

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