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Numerade Educator



Problem 31 Medium Difficulty

(a) Use the Quotient Rule to differentiate the function
$ f(x) = \frac {\tan x - 1}{\sec x} $
(b) Simplify the expression for $ f(x) $ by writing it in terms of $ \sin x $ and $ \cos x $ and then find $ f'(x). $
(c) Show that your answers to parts (a) and (b) are equivalent.


a) $f^{\prime}(x)=\frac{1+\tan x}{\sec x}$
b) $f^{\prime}(x)=\cos x+\sin x$
C) The part $(\mathrm{a})$ answer is equivalent to the part $(\mathrm{b})$ answer.

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Video Transcript

it's clear. So when you read here, so we're gonna use the quotient role we get seek it turns the derivative of tangent minus one minus tangent, minus one times the derivative of Seek it all over secrets where And then we get this to the equal to seek in times sequence square plus zero minus 10 times. Seek it. Times 10 all over. Seek and square Using our trig identities. We're gonna expand this so we get a second, learns vengeance Square, plus seek it minus seek it. Times turn gents Square was seeking time. 10 gin all over Secret Square in this equals one plus two engines over sequined per p. We have Earth of X is equal to 10 gin minus one divided by seeking. So using our trig identities, this is equal to sign over. Co sign minus one all over one over coastline. We're gonna multiply out the dominators. We got signed minus coastline. Next, we're gonna differentiate in respect to X, so we're gonna differentiate these separately. We get derivative of Sinus co sign the derivative of who signed is negative sign. We get co sign plus sign report. See, we're gonna do a we're gonna take our answer, which is one plus tangent over Seek in our first derivative. And we're gonna multiply by co sign for the top and bottom. And when we simplify, we get co sign plus sign. So we proved it.