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differentiation of trig functions (1 Viewer)

Thread starter jellybelly59
Start date Feb 15, 2009

jellybelly59

Active Member

Feb 15, 2009

Show that dy/dx of 1/3 [tanx]^3 - tan x + x = [tanx]^4

$y = \frac{\tan^{3} x}{3} - \tan x + x\\\frac{dy}{dx} = \tan^{2}x(\sec^{2} x) - \sec^{2} x + 1\\= \tan^{2}x(1+\tan^{2}x) - (1+\tan^{2} x) + 1\\= \tan^{2} x + \tan^{4}x - 1 - \tan^{2} x + 1\\\Rightarrow \tan^{4} x \text{ as required.}$