# How do I find the derivative of y=ln(secx + tanx)?

You may use the Chain Rule; you start deriving the logarithm as a normal one and then multipy by the derivative of its argument (which basically is $\frac{1}{\cos} \left(x\right) + \sin \frac{x}{\cos} \left(x\right)$):
$y ' = \frac{1}{\sec \left(x\right) + \tan \left(x\right)} \cdot \left[\sin \frac{x}{\cos} ^ 2 \left(x\right) + \frac{{\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right)}{\cos} ^ 2 \left(x\right)\right] =$
$= \frac{1}{\sec \left(x\right) + \tan \left(x\right)} \cdot \frac{1 + \sin \left(x\right)}{{\cos}^{2} \left(x\right)} = \frac{1}{\cos} \left(x\right)$