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Posts: 24713
9 months ago
 Finding the derivative of inverse tangent 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 $$4$$:$$4y=\tan ^{-1}\left(\frac{4\sin x}{3+5\cos x}\right)$$That tangent of both sides:$$\tan \left(4y\right)=\tan \left(\tan ^{-1}\left(\frac{4\sin x}{3+5\cos x}\right)\right)$$This becomes:$$\tan \left(4y\right)=\frac{4\sin x}{3+5\cos x}$$Now differentiate implicitly with respect to $$x$$. And recall that: $$\frac{d}{dx}\tan u=\sec ^2u\cdot \frac{du}{dx}$$Applying this to the left side ...$$\sec ^2\left(4y\right)\cdot 4\ \frac{dy}{dx}=\frac{4\sin x\left(-5\sin x\right)-4\cos x\left(3+5\cos x\right)}{\left(3+5\cos x\right)^2}$$Solve for $$\frac{dy}{dx}$$:$$\frac{dy}{dx}=\frac{4\sin x\left(-5\sin x\right)-4\cos x\left(3+5\cos x\right)}{4\ \sec ^2\left(4y\right)\left(3+5\cos x\right)^2}$$Simplify the numerator:$$\frac{dy}{dx}=\frac{-20\sin ^2x-12\cos x-20\cos ^2x}{4\ \sec ^2\left(4y\right)\left(3+5\cos x\right)^2}$$Simplify more:$$\frac{dy}{dx}=\frac{-20\left(\sin ^2x+\cos ^2\right)-12\cos x}{4\ \sec ^2\left(4y\right)\left(3+5\cos x\right)^2}$$Recall that: $$\sin ^2x+\cos ^2x=1,\ \therefore$$$$\frac{dy}{dx}=\frac{-5-3\cos x}{\sec ^2\left(4y\right)\left(3+5\cos x\right)^2}$$Simplify more:$$\frac{dy}{dx}=\frac{-1}{\sec ^2\left(4y\right)\left(3+5\cos x\right)}$$There you go! Read 210 times
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