Final answer:
The derivative of y=ln|secx+tanx| is found using the chain rule and the derivatives of the natural logarithm and trigonometric functions, resulting in dy/dx = (1 / |secx + tanx|) * (secx*tanx + sec^2(x)).
Step-by-step explanation:
To find the derivative (dy)/(dx) of the function y=ln|secx+tanx|, we will use the chain rule in combination with the derivatives of the natural logarithm function and trigonometric identities.
The chain rule states that if y = f(u) and u = g(x), then the derivative dy/dx can be found by multiplying the derivative f'(u) by the derivative g'(x), symbolically dy/dx = f'(u) * g'(x). The derivative of ln|u| with respect to u is 1/u when u > 0 and -1/u when u < 0. Since |u| is always positive, we can simply use 1/u for the derivative of the natural logarithm function with respect to |u|.
For the trigonometric part, we know from identities that the derivative of secx is secx*tanx and the derivative of tanx is sec^2(x). Thus, applying the chain rule, we get:
dy/dx = (1 / |secx + tanx|) * (secx*tanx + sec^2(x))
This equation provides the rate of change of y with respect to x for y=ln|secx+tanx|.