Answer:
(a) sin(x) + cos(x)
(b) sin(x) + cos(x)
(c) Both answers are equivalent
Step-by-step explanation:
(a) The given function is:
[tex]f(x) = \frac{tan(x)-1}{sec(x)}[/tex]
According to the quotient rule:
[tex]f(x) = \frac{g(x)}{h(x)}\\f'(x)= \frac{g'(x)h(x)-g(x)h'(x)}{h(x)^2}[/tex]
Applying the quotient rule:
[tex]f(x) = \frac{tan(x)-1}{sec(x)}\\f'(x)=\frac{sec^2(x)*sec(x)-(tan(x)-1)*sec(x)tan(x)}{sec(x)^2}\\f'(x)=\frac{sec^3(x)-sec(x)tan^2(x)+sec(x)tan(x)}{sec(x)^2}\\f'(x)=sec(x)+\frac{tan(x)-tan^2(x)}{sec(x)}\\ \frac{tan(x)}{sec(x)}=sin(x) \\f'(x)=sec(x)+sin(x)-sin(x)tan(x)\\[/tex]
This can be simplified to:
[tex]f'(x)=sec(x)+sin(x)-sin(x)tan(x)\\f'(x) = \frac{1}{cos(x)}+sin(x)-\frac{sin^2(x)}{cos(x)}\\f'(x)=\frac{1+sin(x)cos(x)-sin^2(x)}{cos(x)}\\f'(x)=\frac{sin^2(x)+cos^2(x)+sin(x)cos(x)-sin^2(x)}{cos(x)}\\f'(x)=\frac{cos^2(x)+sin(x)cos(x)}{cos(x)}\\ f'(x)=sin(x) +cos(x)[/tex]
(b) Simplifying in terms of sin(x) and cos(x):
[tex]f(x) = \frac{tan(x)-1}{sec(x)}\\f(x)=\frac{\frac{sin(x)}{cos(x)}-1 }{\frac{1}{cos(x)} } \\f(x)=sin(x)-cos(x)\\f'(x) = cos(x)+sin(x)[/tex]
(c) As proven above, both answers are equivalent.
