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Question Number 133610 by benjo_mathlover last updated on 23/Feb/21
Given f(x) = ∣tan x∣ , find    ((df(x))/dx)∣_(x=k)  where (π/2)<k<π
$$\mathrm{Given}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mid\mathrm{tan}\:\mathrm{x}\mid\:,\:\mathrm{find}\: \\ $$$$\:\frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}\mid_{\mathrm{x}=\mathrm{k}} \:\mathrm{where}\:\frac{\pi}{\mathrm{2}}<\mathrm{k}<\pi \\ $$
Answered by guyyy last updated on 23/Feb/21
Answered by liberty last updated on 23/Feb/21
⇒ we have y^2  = tan^2 x   ⇒2y y′ = 2tan x sec^2 x   ⇒y^′  = ((tan x sec^2 x)/(∣tan x∣)) ; where on interval  (π/2)<k<π ⇒tan k < 0    ((df(x))/dx)∣_(x=k)  = ((tan k sec^2 k)/(−tan k)) = −sec^2  k
$$\Rightarrow\:\mathrm{we}\:\mathrm{have}\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{tan}\:^{\mathrm{2}} \mathrm{x} \\ $$$$\:\Rightarrow\mathrm{2y}\:\mathrm{y}'\:=\:\mathrm{2tan}\:\mathrm{x}\:\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}\: \\ $$$$\Rightarrow\mathrm{y}^{'} \:=\:\frac{\mathrm{tan}\:\mathrm{x}\:\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}}{\mid\mathrm{tan}\:\mathrm{x}\mid}\:;\:\mathrm{where}\:\mathrm{on}\:\mathrm{interval} \\ $$$$\frac{\pi}{\mathrm{2}}<\mathrm{k}<\pi\:\Rightarrow\mathrm{tan}\:\mathrm{k}\:<\:\mathrm{0}\: \\ $$$$\:\frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}\mid_{\mathrm{x}=\mathrm{k}} \:=\:\frac{\mathrm{tan}\:\mathrm{k}\:\mathrm{sec}\:^{\mathrm{2}} \mathrm{k}}{−\mathrm{tan}\:\mathrm{k}}\:=\:−\mathrm{sec}\:^{\mathrm{2}} \:\mathrm{k}\: \\ $$

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