Given: [tex]f(x)=\frac{3}{x}[/tex]
Determine [tex]f^{\prime}(x)[/tex] from first principles.
Answer (1)
Define the derivative from first principles: f ′ ( x ) = lim h → 0 h f ( x + h ) − f ( x ) .
Substitute f ( x ) = x 3 into the definition: f ′ ( x ) = lim h → 0 h x + h 3 − x 3 .
Simplify the expression and cancel h : f ′ ( x ) = lim h → 0 x ( x + h ) − 3 .
Evaluate the limit as h approaches 0: f ′ ( x ) = x 2 − 3 .
Explanation
Problem Introduction We are given the function f ( x ) = x 3 and we need to find its derivative f ′ ( x ) using the definition of the derivative (first principles).
Derivative Definition The definition of the derivative from first principles is: f ′ ( x ) = h → 0 lim h f ( x + h ) − f ( x )
Substitution Substitute f ( x ) = x 3 into the definition: f ′ ( x ) = h → 0 lim h x + h 3 − x 3
Simplification Simplify the expression inside the limit. First, find a common denominator for the fractions in the numerator: x + h 3 − x 3 = x ( x + h ) 3 x − 3 ( x + h ) = x ( x + h ) 3 x − 3 x − 3 h = x ( x + h ) − 3 h
Substitution into Limit Substitute the simplified expression back into the limit: f ′ ( x ) = h → 0 lim h x ( x + h ) − 3 h = h → 0 lim h x ( x + h ) − 3 h
Cancellation Cancel h from the numerator and denominator: f ′ ( x ) = h → 0 lim x ( x + h ) − 3
Evaluation of Limit Evaluate the limit as h approaches 0: f ′ ( x ) = x ( x + 0 ) − 3 = x 2 − 3
Final Answer Therefore, the derivative of f ( x ) = x 3 is: f ′ ( x ) = x 2 − 3
Examples
Understanding derivatives from first principles is crucial in physics, especially when dealing with velocity and acceleration. For instance, if f ( x ) represents the position of an object at time x , then f ′ ( x ) gives the object's velocity. In this case, if the position of an object is given by f ( x ) = x 3 , then the velocity of the object at time x is f ′ ( x ) = x 2 − 3 . This shows how calculus helps us understand motion and change in real-world scenarios.
