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Find f'(x) for each function
f(x)=[tex]\frac{sec(x)}{x}[/tex]
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3.5 Question 179
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Re: 3.5 Question 179
by Eigenvalue » Thu Sep 24, 2026 7:37 pm
f(x)=sec(x)*[tex]\frac{1}{x}[/tex]
=sec(x)*x[tex]x^{-1 }[/tex]
f'(x)=(sec(x))'x[tex]x^{-1}[/tex]+sec(x)([tex]x^{-1 }[/tex])'
=sec(x)tan(x)[tex]x^{-1 }[/tex]+sec(x)(-[tex]x^{-2 }[/tex])
=[tex]\frac{sec(x)tan(x)}{x}[/tex]-[tex]\frac{sec(x)}{x²}[/tex]
=sec(x)*x[tex]x^{-1 }[/tex]
f'(x)=(sec(x))'x[tex]x^{-1}[/tex]+sec(x)([tex]x^{-1 }[/tex])'
=sec(x)tan(x)[tex]x^{-1 }[/tex]+sec(x)(-[tex]x^{-2 }[/tex])
=[tex]\frac{sec(x)tan(x)}{x}[/tex]-[tex]\frac{sec(x)}{x²}[/tex]
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Re: 3.5 Question 179
by nycmath » Sun Sep 27, 2026 3:45 pm
Eigenvalue wrote:f(x)=sec(x)*[tex]\frac{1}{x}[/tex]
=sec(x)*x[tex]x^{-1 }[/tex]
f'(x)=(sec(x))'x[tex]x^{-1}[/tex]+sec(x)([tex]x^{-1 }[/tex])'
=sec(x)tan(x)[tex]x^{-1 }[/tex]+sec(x)(-[tex]x^{-2 }[/tex])
=[tex]\frac{sec(x)tan(x)}{x}[/tex]-[tex]\frac{sec(x)}{x²}[/tex]
By the time we reach Calculus, you will be sick of this stuff.
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