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Deriving the derivative of $\tan^{2}x$ by quotient rule using $$\frac{\sin x}{\cos x}$$ identity, i am getting a different value $2\sec x[\tan x+\tan^{2}x]$ than by directly getting chain rule is $2\sec^{2}x\tan x$

Whats wrong?

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Here is a correct derivation of the derivative of $\tan^2 x$ using the quotient rule:

$$\dfrac d {dx} \tan^2 x = \dfrac d {dx} \dfrac {\sin^2 x}{\cos^2x }=\dfrac{2\sin x \cos x \cos^2x + \sin^2 x 2 \cos x \sin x}{\cos^4x}$$

$$=2\dfrac{\sin x(\cos^2x+\sin^2x)}{\cos^3x}=2\dfrac{\sin x}{\cos^3 x}=2\tan x \sec^2x$$

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  • $\begingroup$ This is equivalent to 2sec^2(x)tan(x) you get by chain rule? And how does 1 check? $\endgroup$ – RandomGuy374 Oct 11 '19 at 6:49
  • $\begingroup$ Yes, the chain rule yields $2 \tan x \sec^2x$ too $\endgroup$ – J. W. Tanner Oct 11 '19 at 6:58
  • $\begingroup$ I forget, how does 1 derive that particular identity?... (cos^2x+sin^2x=1) $\endgroup$ – RandomGuy374 Oct 11 '19 at 7:00
  • $\begingroup$ $\cos^2 x+\sin^2 x = 1$ is the Pythagorean identity $\endgroup$ – J. W. Tanner Oct 11 '19 at 7:08

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