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Find the derivatives of the given functions.$$y=\frac{2 \cos 4 x}{1+\cot 3 x}$$

$\frac{d y}{d x}=\frac{-8 \sin 4 x(1+\cot 3 x)+6 \cos 4 x \csc ^{2} 3 x}{(1+\cot 3 x)^{2}}$

Calculus 1 / AB

Chapter 27

Differentiation of Transcendental Functions

Section 2

Derivatives of the Other Trigonometric Functions

Derivatives

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we want to find the derivative Dy dx with a function Y equals to co sign for X over one plus CO two engine three X two. So if we're going to rely on the derivative shortcuts, we picked up through a single variable calculus and particularly going to use our recently learned trigonometry derivatives to solve. So let's note the kinetic derivatives and other shortcuts. These rules will sit here. I have also extra derivatives here in particular. We need to use DDX, cossacks and D. D. Expo tangent X. We also need to use the chain will and product rule. Specifically if we rewrite Y as to coast four X times one plus CO two into three X men first, we can start this derivative with the product rule and proceed so, using the product rule on co sign for X and one plus CO two into three X negative. First we obtain derivative. Dy dx equals negative to sign for X times four times one plus tangent X negative first my s to co sign for X times one plus tangent three X. A negative second times negative. Costa can't square three X times three. Thus combining like denominators, we have final solution to the bottom two times negative. Four signed for x minus four. Signed for extended three X plus three. Co sign for ex coast. You can square three X all in parenthesis, in the numerator. All over one plus tangent three X square

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