Introduction
In calculus, the derivative of a function represents the rate of change of the function with respect to one of its variables. It is a fundamental concept in mathematics and has numerous applications in various fields, including physics, engineering, and economics. In this article, we will focus on calculating the derivative of a given function and finding its values at specific points.
The Function
The given function is:
p ( ΞΈ ) = 5 ΞΈ { p(\theta) = \sqrt{5 \theta} } p ( ΞΈ ) = 5 ΞΈ β
This function represents a square root function with a coefficient of 5 and a variable ΞΈ.
Calculating the Derivative
To calculate the derivative of the function, we will use the power rule of differentiation, which states that if f(x) = x^n, then f'(x) = nx^(n-1).
In this case, we can rewrite the function as:
p ( ΞΈ ) = ( 5 ΞΈ ) 1 2 { p(\theta) = (5 \theta)^{\frac{1}{2}} } p ( ΞΈ ) = ( 5 ΞΈ ) 2 1 β
Using the power rule, we can differentiate the function as follows:
p β² ( ΞΈ ) = 1 2 ( 5 ΞΈ ) β 1 2 β
5 { p^{\prime}(\theta) = \frac{1}{2} (5 \theta)^{-\frac{1}{2}} \cdot 5 } p β² ( ΞΈ ) = 2 1 β ( 5 ΞΈ ) β 2 1 β β
5
Simplifying the expression, we get:
p β² ( ΞΈ ) = 5 2 5 ΞΈ { p^{\prime}(\theta) = \frac{5}{2 \sqrt{5 \theta}} } p β² ( ΞΈ ) = 2 5 ΞΈ β 5 β
Finding the Values of the Derivative
Now that we have the derivative of the function, we can find its values at specific points.
Finding the Value of the Derivative at ΞΈ = 1
To find the value of the derivative at ΞΈ = 1, we substitute ΞΈ = 1 into the derivative:
p β² ( 1 ) = 5 2 5 β
1 { p^{\prime}(1) = \frac{5}{2 \sqrt{5 \cdot 1}} } p β² ( 1 ) = 2 5 β
1 β 5 β
Simplifying the expression, we get:
p β² ( 1 ) = 5 2 5 { p^{\prime}(1) = \frac{5}{2 \sqrt{5}} } p β² ( 1 ) = 2 5 β 5 β
p β² ( 1 ) = 5 2 β
5 1 2 { p^{\prime}(1) = \frac{5}{2 \cdot 5^{\frac{1}{2}}} } p β² ( 1 ) = 2 β
5 2 1 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 1 2 { p^{\prime}(1) = \frac{5}{2 \cdot 5^{\frac{1}{2}}} } p β² ( 1 ) = 2 β
5 2 1 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
p β² ( 1 ) = 5 2 β
5 { p^{\prime}(1) = \frac{5}{2 \cdot \sqrt{5}} } p β² ( 1 ) = 2 β
5 β 5 β
[ p^{\prime}(1