Question

hw14 the chain rule (target c4; §3.6) score: 5/11 answered: 5/11 question 6 let ( f(x)=2csc(3x)) ( f(x)=) enter an algebraic expression more.

Explanation:

Step1: Recall the chain - rule formula

The chain - rule states that if \(y = f(u)\) and \(u = g(x)\), then \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\). Let \(u = 3x\), so \(y = 2\csc(u)\).

Step2: Find the derivative of the outer function

The derivative of \(y = 2\csc(u)\) with respect to \(u\) is \(\frac{dy}{du}=- 2\csc(u)\cot(u)\).

Step3: Find the derivative of the inner function

The derivative of \(u = 3x\) with respect to \(x\) is \(\frac{du}{dx}=3\).

Step4: Apply the chain - rule

By the chain - rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\). Substitute \(\frac{dy}{du}=-2\csc(u)\cot(u)\) and \(\frac{du}{dx}=3\) into the formula. Replace \(u\) with \(3x\), we get \(f^\prime(x)=-6\csc(3x)\cot(3x)\).

Answer:

\(-6\csc(3x)\cot(3x)\)