Find [tex]\frac{dy}{dx}[/tex] if (a) [tex]$y=6 x^2 \sin x \cos x$[/tex]
Answer (1)
Rewrite the function using the identity 2 sin x cos x = sin ( 2 x ) to get y = 3 x 2 sin ( 2 x ) .
Apply the product rule: d x d y = u ′ v + u v ′ , where u = 3 x 2 and v = sin ( 2 x ) .
Find the derivatives: u ′ = 6 x and v ′ = 2 cos ( 2 x ) .
Substitute and simplify: d x d y = 6 x sin ( 2 x ) + 6 x 2 cos ( 2 x ) = 6 x ( x cos ( 2 x ) + sin ( 2 x )) .
Explanation
Problem Analysis We are given the function y = 6 x 2 sin x cos x and we need to find its derivative with respect to x , denoted as d x d y . This problem involves finding the derivative of a product of functions, so we will use the product rule.
Simplifying the Function First, let's rewrite the function using the trigonometric identity 2 sin x cos x = sin ( 2 x ) . So, y = 6 x 2 sin x cos x = 3 x 2 ( 2 sin x cos x ) = 3 x 2 sin ( 2 x ) . This simplifies the differentiation process.
Applying the Product Rule Now, we apply the product rule to differentiate y = 3 x 2 sin ( 2 x ) . The product rule states that if y = u ( x ) v ( x ) , then d x d y = u ′ ( x ) v ( x ) + u ( x ) v ′ ( x ) . In our case, u ( x ) = 3 x 2 and v ( x ) = sin ( 2 x ) .
Finding the Derivatives We need to find the derivatives of u ( x ) and v ( x ) .
d x d ( 3 x 2 ) = 6 x
d x d ( sin ( 2 x )) = 2 cos ( 2 x )
Substituting into the Product Rule Now, we substitute these derivatives back into the product rule formula:
d x d y = ( 6 x ) ( sin ( 2 x )) + ( 3 x 2 ) ( 2 cos ( 2 x ))
d x d y = 6 x sin ( 2 x ) + 6 x 2 cos ( 2 x )
Simplifying the Expression We can factor out 6 x from the expression:
d x d y = 6 x ( sin ( 2 x ) + x cos ( 2 x ))
Final Answer Thus, the derivative of y = 6 x 2 sin x cos x with respect to x is d x d y = 6 x ( x cos ( 2 x ) + sin ( 2 x )) .
Examples
Understanding derivatives is crucial in physics, especially when analyzing motion. For example, if x ( t ) represents the position of an object at time t , then its velocity v ( t ) is the derivative of x ( t ) with respect to t , i.e., v ( t ) = d t d x . Similarly, the acceleration a ( t ) is the derivative of the velocity with respect to time, a ( t ) = d t d v . Derivatives help us understand how quantities change over time, which is fundamental in studying dynamics and kinematics.
