If $f(x)=5 \sqrt{x}(x^3-3 \sqrt{x}+7)$, find $f^{\prime}(x)$.
Find $f^{\prime}(1)$.
Answer (1)
Rewrite the function: f ( x ) = 5 x 7/2 − 15 x + 35 x 1/2 .
Find the derivative using the power rule: f ′ ( x ) = ( 35/2 ) x 5/2 − 15 + ( 35/2 ) x − 1/2 .
Evaluate the derivative at x = 1 : f ′ ( 1 ) = ( 35/2 ) ( 1 ) 5/2 − 15 + ( 35/2 ) ( 1 ) − 1/2 .
Simplify to find the final answer: 20 .
Explanation
Rewrite the function We are given the function f ( x ) = 5 x ( x 3 − 3 x + 7 ) and we need to find its derivative f ′ ( x ) and then evaluate f ′ ( 1 ) . First, let's rewrite the function to make it easier to differentiate:
f ( x ) = 5 x 1/2 ( x 3 − 3 x 1/2 + 7 ) = 5 x 7/2 − 15 x + 35 x 1/2
Find the derivative Now, we find the derivative f ′ ( x ) using the power rule, which states that if g ( x ) = a x n , then g ′ ( x ) = na x n − 1 . Applying the power rule to each term in f ( x ) , we get:
f ′ ( x ) = 5 ⋅ ( 7/2 ) ⋅ x ( 7/2 − 1 ) − 15 ⋅ 1 ⋅ x ( 1 − 1 ) + 35 ⋅ ( 1/2 ) ⋅ x ( 1/2 − 1 )
Simplifying the exponents, we have:
f ′ ( x ) = ( 35/2 ) x 5/2 − 15 + ( 35/2 ) x − 1/2
Evaluate the derivative at x=1 Next, we evaluate f ′ ( 1 ) by substituting x = 1 into the expression for f ′ ( x ) :
f ′ ( 1 ) = ( 35/2 ) ( 1 ) 5/2 − 15 + ( 35/2 ) ( 1 ) − 1/2 = ( 35/2 ) − 15 + ( 35/2 ) = 35 − 15 = 20
Final Answer Therefore, f ′ ( 1 ) = 20 .
Examples
Understanding derivatives is crucial in many real-world applications. For example, if f ( x ) represents the position of a car at time x , then f ′ ( x ) represents the velocity of the car at time x . Evaluating f ′ ( 1 ) would give the car's velocity at time x = 1 . This concept extends to various fields like physics, engineering, and economics, where understanding rates of change is essential for modeling and optimization.
