How do you find the derivative of # y=ln|secx+tanx|#?

Question

Christopher Agresta · Accepted Answer

# \qquad \qquad \qquad \qquad \qquad \qquad y' \ = \ secx. # # "We will do this by the Chain Rule." # # "Recall:" \qquad \qquad \qquad y \ = \ ln | x | \qquad rArr \qquad y' \ = \ 1/x. # # "So, by the Chain Rule:" # # \qquad \quad y \ = \ ln | f(x) | \qquad rArr \qquad y' \ = \ [ 1/f(x) ] cdot f'(x) \ = \ { f'(x) }/f(x). # # \qquad :. \qquad \qquad y \ = \ ln | f(x) | \qquad rArr \qquad y' \ = \ { f'(x) }/f(x). # # "So, in our example:" # # \qquad \qquad \qquad \qquad \qquad y \ = \ ln| secx + tanx |; \qquad \qquad \quad \ \color{blue}{ f(x) \ = \ secx + tanx } # # \qquad \qquad \qquad \qquad \quad y' \ = \ { ( secx + tanx )' }/{ secx + tanx }; \qquad \qquad \quad \color{blue}{ = { \ f'(x) }/f(x) } # # \qquad \qquad \qquad \qquad \qquad \quad \ = \ { ( secx )'+ ( tanx )' }/{ secx + tanx }; # # \qquad \qquad \qquad \qquad \qquad \quad \ = \ { secx tanx+ sec^2x }/{ secx + tanx }; # # \qquad \qquad \qquad \qquad \qquad \quad \ = \ { secx ( tanx+ secx ) }/{ secx + tanx }; # # \qquad \qquad \qquad \qquad \qquad \quad \ = \ { secx ( secx + tanx ) }/{ secx + tanx }; # # \qquad \qquad \qquad \qquad \qquad \quad \ = \ { secx color{red}cancel{ ( secx + tanx ) } }/color{red}cancel{ ( secx + tanx ) }; # # \qquad \qquad \qquad \qquad \qquad \quad \ = \ secx. # # "So, we have now:" # # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad y' \ = \ secx. # # "Summarizing:" # # \qquad \qquad \qquad \qquad y \ = \ ln| secx + tanx | \qquad rArr \qquad y' \ = \ secx. #

Samantha Adkison · Answer

undefined To find the derivative of $ y = \ln|\sec(x) + \tan(x)| $, we use the chain rule and the derivative of the absolute value function.

First, note that $ \sec(x) = \frac{1}{\cos(x)} $ and $ \tan(x) = \frac{\sin(x)}{\cos(x)} $.

Let $ u = \sec(x) + \tan(x) $.

Then, $ \ln|u| = \ln(u) $ because $ u $ is always positive for any real $ x $ (since $ \sec(x) $ and $ \tan(x) $ are positive in the given domain).

Now, we find $ \frac{dy}{dx} $:

$ \frac{dy}{dx} = \frac{d}{dx}[\ln(u)] $

Using the chain rule, this becomes:

$ \frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} $

Now, find $ \frac{du}{dx} $ using the derivatives of $ \sec(x) $ and $ \tan(x) $:

$ \frac{du}{dx} = \frac{d}{dx}[\sec(x) + \tan(x)] $

$ = \sec(x)\tan(x) + \sec^2(x) $

Now, substitute $ u = \sec(x) + \tan(x) $ and $ \frac{du}{dx} = \sec(x)\tan(x) + \sec^2(x) $ back into $ \frac{dy}{dx} $:

$ \frac{dy}{dx} = \frac{1}{\sec(x) + \tan(x)} \cdot [\sec(x)\tan(x) + \sec^2(x)] $

Simplify:

$ \frac{dy}{dx} = \sec(x)\tan(x) + \sec^2(x) $

Therefore, the derivative of $ y = \ln|\sec(x) + \tan(x)| $ is $ \sec(x)\tan(x) + \sec^2(x) $.