Use the chain rule to solve: [tex]y=e^x(3 x-1)^4[/tex]
Answer (1)
Apply the product rule: d x d y = u ′ v + u v ′ , where u = e x and v = ( 3 x − 1 ) 4 .
Find the derivatives: u ′ = e x and v ′ = 4 ( 3 x − 1 ) 3 ⋅ 3 = 12 ( 3 x − 1 ) 3 using the chain rule.
Substitute the derivatives into the product rule: d x d y = e x ( 3 x − 1 ) 4 + e x [ 12 ( 3 x − 1 ) 3 ] .
Simplify the expression: d x d y = e x ( 3 x − 1 ) 3 ( 3 x + 11 ) .
d x d y = e x ( 3 x − 1 ) 3 ( 3 x + 11 )
Explanation
Problem Analysis We are given the function y = e x ( 3 x − 1 ) 4 and asked to find its derivative using the chain rule. This problem requires us to apply both the product rule and the chain rule.
Applying the Product Rule First, we apply the product rule, which 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 ) = e x and v ( x ) = ( 3 x − 1 ) 4 .
Derivative of u(x) Now we find the derivatives of u ( x ) and v ( x ) . The derivative of u ( x ) = e x is simply u ′ ( x ) = e x .
Derivative of v(x) To find the derivative of v ( x ) = ( 3 x − 1 ) 4 , we use the chain rule. The chain rule states that if v ( x ) = [ f ( x ) ] n , then v ′ ( x ) = n [ f ( x ) ] n − 1 ⋅ f ′ ( x ) . Here, f ( x ) = 3 x − 1 and n = 4 . So, v ′ ( x ) = 4 ( 3 x − 1 ) 4 − 1 ⋅ d x d ( 3 x − 1 ) = 4 ( 3 x − 1 ) 3 ⋅ 3 = 12 ( 3 x − 1 ) 3 .
Substituting into the Product Rule Now we substitute u ′ ( x ) and v ′ ( x ) back into the product rule formula: d x d y = u ′ ( x ) v ( x ) + u ( x ) v ′ ( x ) = e x ( 3 x − 1 ) 4 + e x [ 12 ( 3 x − 1 ) 3 ] .
Simplifying the Expression We can simplify this expression by factoring out e x ( 3 x − 1 ) 3 : d x d y = e x ( 3 x − 1 ) 3 [( 3 x − 1 ) + 12 ] = e x ( 3 x − 1 ) 3 ( 3 x − 1 + 12 ) = e x ( 3 x − 1 ) 3 ( 3 x + 11 ) .
Final Answer Therefore, the derivative of y = e x ( 3 x − 1 ) 4 is d x d y = e x ( 3 x − 1 ) 3 ( 3 x + 11 ) .
Examples
In physics, if you're analyzing the motion of a particle where its position is given by a function like y = e x ( 3 x − 1 ) 4 , finding the derivative helps determine the particle's velocity. The product and chain rules are essential for understanding how different factors contribute to the overall rate of change, providing insights into the particle's dynamics.
