Question

1. $f(x)=3sin 2x$

t. $6cos (2x)$
b. $3cos (2x)$
f. $6sin (2x)$
k. $3sin (2x)$

Explanation:

Step1: Apply the chain rule

The chain rule states that if \(y = f(g(x))\), then \(y^\prime=f^\prime(g(x))\cdot g^\prime(x)\). For \(y = 3\sin(2x)\), let \(u = 2x\), so \(y = 3\sin(u)\). The derivative of \(\sin(u)\) with respect to \(u\) is \(\cos(u)\), and the derivative of \(u = 2x\) with respect to \(x\) is \(2\).

Step2: Calculate the derivative

Using the chain rule \(y^\prime=(3\cos(u))\cdot2\). Substituting back \(u = 2x\), we get \(y^\prime = 6\cos(2x)\)

Answer:

T. \(6\cos(2x)\)