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Question Number 7900 by tawakalitu last updated on 23/Sep/16
Find an equation of the tangent line to the curve   y = tan^2 x   at  the point  ((π/3), 3)
$${Find}\:{an}\:{equation}\:{of}\:{the}\:{tangent}\:{line}\:{to}\:{the}\:{curve}\: \\ $$$${y}\:=\:{tan}^{\mathrm{2}} {x}\:\:\:{at}\:\:{the}\:{point}\:\:\left(\frac{\pi}{\mathrm{3}},\:\mathrm{3}\right) \\ $$
Commented by sou1618 last updated on 24/Sep/16
l:tangent line  3=tan^2 ((π/3))  so point(π/3,3) is on l    y ′=2tan(x)∙(tan x)′  y ′=2tan(x)∙(1/(cos^2 x))  l:y−3=2tan(π/3)∙(1/(cos^2 (π/3)))(x−π/3)  l:y=2(√3)∙4(x−π/3)+3  l:y=8(√3)x+3−((8π(√3))/3)  l:y=8(√3)x+3−((8π)/( (√3))) .
$${l}:{tangent}\:{line} \\ $$$$\mathrm{3}={tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{3}}\right) \\ $$$${so}\:{point}\left(\pi/\mathrm{3},\mathrm{3}\right)\:{is}\:{on}\:{l} \\ $$$$ \\ $$$${y}\:'=\mathrm{2}{tan}\left({x}\right)\centerdot\left({tan}\:{x}\right)' \\ $$$${y}\:'=\mathrm{2}{tan}\left({x}\right)\centerdot\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}} \\ $$$${l}:{y}−\mathrm{3}=\mathrm{2}{tan}\left(\pi/\mathrm{3}\right)\centerdot\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left(\pi/\mathrm{3}\right)}\left({x}−\pi/\mathrm{3}\right) \\ $$$${l}:{y}=\mathrm{2}\sqrt{\mathrm{3}}\centerdot\mathrm{4}\left({x}−\pi/\mathrm{3}\right)+\mathrm{3} \\ $$$${l}:{y}=\mathrm{8}\sqrt{\mathrm{3}}{x}+\mathrm{3}−\frac{\mathrm{8}\pi\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${l}:{y}=\mathrm{8}\sqrt{\mathrm{3}}{x}+\mathrm{3}−\frac{\mathrm{8}\pi}{\:\sqrt{\mathrm{3}}}\:. \\ $$$$ \\ $$
Commented by tawakalitu last updated on 24/Sep/16
Thank you so much.
$${Thank}\:{you}\:{so}\:{much}.\: \\ $$
Answered by prakash jain last updated on 02/Oct/16
see comments
$$\mathrm{see}\:\mathrm{comments} \\ $$

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