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Title: Find derivative of complicated inverse trig function Post by: bio_man on Oct 2, 2018 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. |