Rewrite the function: f ( x ) = − 2 ( sin ( 3 x 4 ) ) 2 .
Apply the chain rule multiple times.
Simplify the expression using the identity 2 sin ( a ) cos ( a ) = sin ( 2 a ) .
The derivative is: − 24 x 3 sin ( 6 x 4 ) .
Explanation
Problem Analysis We are asked to find the derivative of the function f ( x ) = − 2 sin 2 ( 3 x 4 ) . This requires using the chain rule multiple times.
Rewriting the Function First, rewrite the function as f ( x ) = − 2 ( sin ( 3 x 4 ) ) 2 .
Differentiating the Outer Function Apply the chain rule. Differentiate the outer function first. Let u = sin ( 3 x 4 ) . Then f ( x ) = − 2 u 2 . The derivative of − 2 u 2 with respect to u is − 4 u .
Differentiating the Sine Function Now, differentiate u = sin ( 3 x 4 ) with respect to x . The derivative is cos ( 3 x 4 ) ⋅ d x d ( 3 x 4 ) .
Differentiating the Inner Function Next, differentiate 3 x 4 with respect to x . The derivative is 12 x 3 .
Combining the Derivatives Combine the derivatives using the chain rule: f ′ ( x ) = − 4 sin ( 3 x 4 ) ⋅ cos ( 3 x 4 ) ⋅ 12 x 3 .
Simplifying the Expression Simplify the expression: f ′ ( x ) = − 48 x 3 sin ( 3 x 4 ) cos ( 3 x 4 ) .
Further Simplification Use the trigonometric identity 2 sin ( a ) cos ( a ) = sin ( 2 a ) to further simplify. In our case, a = 3 x 4 , so 2 sin ( 3 x 4 ) cos ( 3 x 4 ) = sin ( 6 x 4 ) . Therefore, f ′ ( x ) = − 24 x 3 sin ( 6 x 4 ) .
Final Answer The derivative of the given function is f ′ ( x ) = − 24 x 3 sin ( 6 x 4 ) .
Examples
In physics, if you have a system where the position of an object is described by a function involving trigonometric functions like the one in this problem, finding the derivative helps you determine the object's velocity. For example, if the angular displacement of a pendulum is given by − 2 sin 2 ( 3 t 4 ) , where t is time, then its angular velocity is − 24 t 3 sin ( 6 t 4 ) . This is crucial for analyzing the pendulum's motion and energy.
The derivative of the function f ( x ) = − 2 sin 2 ( 3 x 4 ) is f ′ ( x ) = − 24 x 3 sin ( 6 x 4 ) . We apply the chain rule multiple times to differentiate the outer and inner functions, simplifying the result using trigonometric identities. This process highlights the power of the chain rule in calculus.
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