How to get the derivative of $(\ln(x))^{\sec(x)}$? How do you get the derivative of $(\ln(x))^{\sec(x)}$? I know that the derivative of $\ln(x)$ is $\frac 1x$ but what happens when you take it to an exponent of $\sec(x)$?
 A: Note that
$$\large(\ln x)^{\sec x}=(e^{\ln\ln x})^{\sec x}=e^{(\ln\ln x)(\sec x)}\;;$$
now differentiate it as you would any exponential, not forgetting to use the chain rule.
A: Let$$y = \ln(x)^{\sec(x)}$$
Now sse implicit differentiation to find the derivative
$$\ln y=\ln \left(\ln(x)^{\sec(x)}\right)$$
$$\ln y=\sec (x)\ln \left(\ln(x)\right)$$
$$\frac{d}{dx}\left(\ln y\right) = \frac{d}{dx} \left(\sec (x)\ln \left(\ln(x)\right)\right)$$
$$\frac 1y \frac{dy}{dx}= \frac{d}{dx} \left(\frac{1}{\cos (x)}\ln \left(\ln(x)\right)\right)$$
$$\frac 1y \frac{dy}{dx}= \frac{d}{dx}\left(\frac{1}{\cos (x)}\right)\left(\ln \left(\ln(x)\right)\right)+\frac{d}{dx}\left(\left(\ln \left(\ln(x)\right)\right)\right)\frac{1}{\cos (x)}$$
$$\frac 1y \frac{dy}{dx}= \sec(x)\tan(x)\left(\ln \left(\ln(x)\right)\right)+ 
\frac 1y \frac{1}{\ln(x)}
\frac{1}{\cos (x)}$$
$$\frac 1y \frac{dy}{dx}= \frac{\sin(x)}{\cos^2(x)}\ln \left(\ln(x)\right)+ 
 \frac{1}{x\ln(x)\cos(x)}$$
$$\frac{dy}{dx}=\ln(x)^{\sec(x)}\left(\frac{\sin(x)\ln(\ln(x))}{\cos^2(x)}+\frac{1}{x\ln(x)\cos(x)}\right)$$
A: Hint: what does exponentiation really means?
$$a^b = \exp (b \ln a)$$
If you can find the derivative of function composition, you are almost done.
Remember, $\sec x = \frac{1}{\cos x}$ and the derivative of $f \circ g$ is $(f' \circ g) \cdot g'$.
And of course, be very carfeul about where your function is defined (logarithm of a negative number must be avoided).
