How do you differentiate $f(x)=(1-secx)/(tanx)$ ?

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Answer 1 · 2017-01-08T14:51:15

$f(x) = (1 - 1/cosx)/(sinx/cosx)$

$f(x) = ((cosx- 1)/cosx)/(sinx/cosx)$

$f(x) = (cosx - 1)/cosx * cosx/sinx$

$f(x) = (cosx - 1)/sinx$

$f'(x) = (-sinx(sinx) - (cosx - 1)(cosx))/(sinx)^2$

$f'(x) = (-sin^2x - cos^2x + cosx)/(sinx)^2$

$f'(x) = (-(sin^2x + cos^2x) + cosx)/sin^2x$

$f'(x) = (cosx - 1)/sin^2x$

Hopefully this helps!

Answer 2 · 2017-01-08T15:39:55

$f'(x)=cscx(cotx-cscx)$ .

$f(x)=(1-secx)/tanx=1/tanx-secx/tanx=cotx-secx/(sinxsecx)$

$:. f(x)=cotx-cscx$

$:. f'(x)=-csc^2x-(-cscxcotx)=cscx(cotx-cscx).$

How do you differentiate f(x)=(1-secx)/(tanx)f(x)=(1-secx)/(tanx)?