Title: Finding the derivative of inverse tangent Post by: bio_man on Oct 2, 2018 From the image, your function is: \(y=\frac{1}{4}\tan ^{-1}\left(\frac{4\sin x}{3+5\cos x}\right)\)
Start by multiplying both sides by \(4\): \(4y=\tan ^{-1}\left(\frac{4\sin x}{3+5\cos x}\right)\) \(\tan \left(4y\right)=\tan \left(\tan ^{-1}\left(\frac{4\sin x}{3+5\cos x}\right)\right)\) \(\tan \left(4y\right)=\frac{4\sin x}{3+5\cos x}\) Applying this to the left side ... \(\sec ^2\left(4y\right)\cdot 4\ \frac{dy}{dx}=\frac{4\sin x\left(-5\sin x\right)-4\cos x\left(3+5\cos x\right)}{\left(3+5\cos x\right)^2}\) \(\frac{dy}{dx}=\frac{4\sin x\left(-5\sin x\right)-4\cos x\left(3+5\cos x\right)}{4\ \sec ^2\left(4y\right)\left(3+5\cos x\right)^2}\) \(\frac{dy}{dx}=\frac{-20\sin ^2x-12\cos x-20\cos ^2x}{4\ \sec ^2\left(4y\right)\left(3+5\cos x\right)^2}\) Simplify more: \(\frac{dy}{dx}=\frac{-20\left(\sin ^2x+\cos ^2\right)-12\cos x}{4\ \sec ^2\left(4y\right)\left(3+5\cos x\right)^2}\) \(\frac{dy}{dx}=\frac{-5-3\cos x}{\sec ^2\left(4y\right)\left(3+5\cos x\right)^2}\) Simplify more: \(\frac{dy}{dx}=\frac{-1}{\sec ^2\left(4y\right)\left(3+5\cos x\right)}\) There you go! ;D Title: Re: Finding the derivative of inverse tangent Post by: duddy on Nov 7, 2019 ☝️👍
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