Problem #1: Differentiate the given function.
(a) [tex]f(x)=x^4 e^{8 x}[/tex]
(b) [tex]f(x)=\ln \left(e^{-4 x}+9 x\right)[/tex]

For f ( x ) = x 4 e 8 x , apply the product rule: f ′ ( x ) = 4 x 3 e 8 x + 8 x 4 e 8 x .
For f ( x ) = ln ( e − 4 x + 9 x ) , apply the chain rule: f ′ ( x ) = e − 4 x + 9 x − 4 e − 4 x + 9 ​ .
The derivative of f ( x ) = x 4 e 8 x is 4 x 3 e 8 x + 8 x 4 e 8 x ​ .
The derivative of f ( x ) = ln ( e − 4 x + 9 x ) is e − 4 x + 9 x − 4 e − 4 x + 9 ​ ​ .

Explanation

Problem Analysis We are asked to find the derivatives of two functions.

Part (a) - Applying the Product Rule (a) We have f ( x ) = x 4 e 8 x . This requires the product rule, which states that ( uv ) ′ = u ′ v + u v ′ , where u = x 4 and v = e 8 x .

Finding Derivatives of u and v We find the derivatives of u and v : u ′ = 4 x 3 and v ′ = 8 e 8 x .

Applying the Product Rule Applying the product rule, we get f ′ ( x ) = ( 4 x 3 ) ( e 8 x ) + ( x 4 ) ( 8 e 8 x ) = 4 x 3 e 8 x + 8 x 4 e 8 x .

Part (b) - Applying the Chain Rule (b) We have f ( x ) = ln ( e − 4 x + 9 x ) . This requires the chain rule, which states that ( ln ( u ) ) ′ = u u ′ ​ , where u = e − 4 x + 9 x .

Finding Derivative of u We find the derivative of u : u ′ = − 4 e − 4 x + 9 .

Applying the Chain Rule Applying the chain rule, we get f ′ ( x ) = e − 4 x + 9 x − 4 e − 4 x + 9 ​ .


Examples
Understanding derivatives is crucial in many real-world applications. For instance, in physics, the derivative of a position function gives the velocity, and the derivative of the velocity gives the acceleration. In economics, derivatives are used to find marginal cost and marginal revenue, which help businesses optimize their production and pricing strategies. In engineering, derivatives are used to analyze the stability and control of systems.

Answered by GinnyAnswer | 2025-07-07