what is the derivative of f(x)=tan(x)cot(x)?

Question

Help clarify please. When I worked this problem I got cot(x)*(sec(x))^2-(cdc(x))^2*tan(x) but when I type it into wolfram alpha it says the derivative is zero. Am I right or are they?

Answer 1

f(x)= tanx cotx

= 1

so f'(x) = 0

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f(x) = tanx cotx

f' (x) = tanx (- cosex^2x) + cotx ( sec^2x)

= - sinx/cosx sin^2x + cosx/sinx* cos^2x

= - 1/sinx cosx + 1/sinx cosx

= 0

so both way the result is same [ zero ]

Answer 2

f(x) = tan(x)cot(x)            ← Simplify to something easier to differentiate

f(x) = (sinx/cosx)(cosx/sinx)

f(x) = 1            ← Now, differentiate



f'(x) = 0            ← ANSWER



Have a good one!

.

Answer 3

You both are. As are Ray S, Rameshwar, and Mathmom. (TU's!)



Except that at all integer multiples of ½π, it is undefined.

Answer 4

(sec^2x )(cotx)+(-csc^2x)(tanx) should be correct.

the zero may be because the two are opposite functions.