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

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Answer 1 · 2018-03-02T10:39:34

# \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. #