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!