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Educator
Posts: 23128
3 months ago
 Find derivative of complicated inverse trig function Derive:$$y=\operatorname{arccot}x+\arctan \left(\frac{2+x}{1-2x}\right)$$Before we start, recall:$$\frac{d}{dx}\operatorname{arccot}x=\frac{-1}{1+x^2}$$$$\frac{d}{dx}\arctan x=\frac{1}{1+x^2}$$Now, let's return to the problem, the left side is easy, it becomes: $$\frac{dy}{dx}$$, but the right side, specifically the tangent part requires the chain rule:$$\frac{dy}{dx}=\frac{-1}{1+x^2}+\frac{1}{1+\left(\frac{2+x}{1-2x}\right)^2}\left(\frac{\left(2+x\right)\left(-2\right)-\left(1-2x\right)}{\left(1-2x\right)^2}\right)$$The derivative is done. Now we clean it:$$\frac{dy}{dx}=\frac{-1}{1+x^2}+\frac{1}{1+\left(\frac{2+x}{1-2x}\right)^2}\left(\frac{-4-2x-1+2x}{\left(1-2x\right)^2}\right)$$Clean more:$$\frac{dy}{dx}=\frac{-1}{1+x^2}+\frac{1}{1+\left(\frac{2+x}{1-2x}\right)^2}\left(\frac{-5}{\left(1-2x\right)^2}\right)$$Clean more:$$\frac{dy}{dx}=\frac{-1}{1+x^2}+\frac{-5}{\left[1+\left(\frac{2+x}{1-2x}\right)^2\right]\left(1-2x\right)^2}$$If you want, you can combine more, but this is fine. Read 153 times
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