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Finding the derivative of inverse tangent
bio_man
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A year 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!
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