Question

tercera derivada de 3tanx

Answer 1

$y=3\tan x \\ y' =3 \cdot (\tan x)' = \frac{3}{\cos^2 x} \\ \\ y'' = \frac{-3(\cos^2 x)'}{\cos^4 x}= \frac{-3 \cdot 2\cos x \cdot (\cos x)'}{\cos^4 x}= \frac{-6 \cos x \cdot (-\sin x)}{\cos^4 x}= \frac{6\cos x \sin x}{\cos^4 x}=\frac{6\sin x}{\cos^3 x}$

$y''' = \frac{(6\sin x)' \cdot \cos^3 x-(\cos^3 x)' \cdot 6\sin x}{\cos^6 x}=\frac{6 \cos x \cdot \cos^3 x-3\cos^2 x \cdot (\cos x)' \cdot 6\sin x}{\cos^6x}=\\ = \frac{6\cos^4 x-3\cos^2 x \cdot (-\sin x) \cdot 6\sin x}{\cos^6 x}= \frac{6\cos^4 x +18\cos^2 x \sin^2 x}{\cos^6x}= \frac{6\cos^2 x+18\sin^2 x}{\cos^4 x}=\\ =\frac{6\cos^2 x +18(1-\cos^2 x)}{\cos^4 x}= \frac{6\cos^2 x +18-18\cos^2 x}{\cos^4 x}=\frac{-12\cos^2 x+18}{\cos^4 x}= \\ = \frac{-12\cos^2 x}{\cos^4 x}+\frac{18}{\cos^4 x}=$
$=-\frac{12}{\cos^2 x}+\frac{18}{\cos^4x}=-12\sec^2 x+18\sec^4 x=\boxed{6\sec^2 x(3\sec^2 x-2)}$