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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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