Given that [tex]y =\sin \left(2 x^3-4 x+2\right)[/tex], find the derivative of y with respect to x.
(a) [tex]2 x^3-4 x \sin \left(2 x^3-4 x+2\right)[/tex]
(b) [tex]6 x^2-4 \cos \left(2 x^3-4 x+2\right)[/tex]
(c) [tex]4 x^2-4 \sin \left(2 x^3-4 x+2\right)[/tex]
(d) [tex]\sin \left(2 x^3-4 x+2\right)[/tex]
Answer (2)
The derivative of y = sin ( 2 x 3 − 4 x + 2 ) is calculated using the chain rule, resulting in d x d y = ( 6 x 2 − 4 ) cos ( 2 x 3 − 4 x + 2 ) , which corresponds to option (b).
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Recognize the need for the chain rule due to the composite function y = sin ( 2 x 3 − 4 x + 2 ) .
Let u = 2 x 3 − 4 x + 2 , transforming the function to y = sin ( u ) , and apply the chain rule: d x d y = d u d y ⋅ d x d u .
Compute d u d y = cos ( u ) and d x d u = 6 x 2 − 4 .
Substitute back to find d x d y = ( 6 x 2 − 4 ) cos ( 2 x 3 − 4 x + 2 ) , which corresponds to option (b). 6 x 2 − 4 cos ( 2 x 3 − 4 x + 2 )
Explanation
Problem Analysis We are given the function y = sin ( 2 x 3 − 4 x + 2 ) and we need to find its derivative with respect to x . This requires us to use the chain rule.
Applying the Chain Rule Let u = 2 x 3 − 4 x + 2 . Then y = sin ( u ) . According to the chain rule, d x d y = d u d y ⋅ d x d u .
Derivative of Sine Function First, we find the derivative of y with respect to u : d u d y = d u d ( sin ( u )) = cos ( u ) .
Derivative of Polynomial Next, we find the derivative of u with respect to x : d x d u = d x d ( 2 x 3 − 4 x + 2 ) = 6 x 2 − 4 .
Substituting Back Now, we substitute these derivatives back into the chain rule formula: d x d y = cos ( u ) ⋅ ( 6 x 2 − 4 ) = ( 6 x 2 − 4 ) cos ( 2 x 3 − 4 x + 2 ) .
Final Answer Comparing our result with the given options, we see that it matches option (b).
Examples
The derivative of a function is a fundamental concept in calculus with numerous real-world applications. For instance, in physics, if y represents the position of an object as a function of time x , then d x d y gives the object's velocity at any given time. Similarly, in economics, if y represents the cost of production as a function of the number of units produced x , then d x d y gives the marginal cost, which is the cost of producing one additional unit. Understanding derivatives allows us to analyze rates of change and optimize various processes.
