Question

Evaluate the derivative of the following functions at the given point. $$c=2 \sqrt{s}-1 ; s=25$$

Short answer

Based on the step by step solution, find the derivative of the function $$c = 2 \sqrt{s} - 1$$ and evaluate it at $$s = 25$$. The derivative of the given function is $$\frac{dc}{ds} = s^{-\frac{1}{2}}$$, and when evaluated at $$s = 25$$, the value of the derivative is $$\frac{1}{5}$$.

Step-by-step solution

Differentiate the given function

First, we will find the derivative of the function $$c = 2 \sqrt{s} - 1$$ with respect to the variable s, which can be expressed as $$\frac{dc}{ds}$$. Using the power rule for differentiation, we have: $$\frac{dc}{ds} = \frac{d}{ds}(2 \sqrt{s} - 1)$$

Rewrite the square root as a power

To apply the differentiation power rule, we need to rewrite the square root as a power. We know that $$\sqrt{s} = s^{\frac{1}{2}}$$. So, the function becomes: $$c = 2 s^{\frac{1}{2}} - 1$$

Apply the power rule for differentiation

Now, we compute the derivative of the function $$c = 2 s^{\frac{1}{2}} - 1$$ with respect to the variable s. Using the power rule, we have: $$\frac{dc}{ds} = \frac{d}{ds}(2 s^{\frac{1}{2}}) - \frac{d}{ds}(1)$$ The derivative of a constant (1 in this case) is zero, so we will only need to differentiate the term with s. $$\frac{dc}{ds} = 2 \cdot \frac{1}{2} s^{\frac{1}{2}-1}$$

Simplify the derivative expression

Now, we simplify the expression for the derivative: $$\frac{dc}{ds} = s^{-\frac{1}{2}}$$

Evaluate the derivative at the given point

Finally, we need to evaluate the derivative $$\frac{dc}{ds} = s^{-\frac{1}{2}}$$ at the given point $$s = 25$$: $$\frac{dc}{ds}\Bigr|_{s=25} = 25^{-\frac{1}{2}} = \frac{1}{5}$$ So, the value of the derivative at the given point is $$\frac{1}{5}$$.

Key concepts

Power Rule

The Power Rule is a basic guideline in calculus that makes finding derivatives straightforward, especially for polynomial functions. In essence, it helps us differentiate functions that are powers of a variable, such as \(x^n\). To apply the power rule:

• Multiply the entire term by the exponent.
• Subtract one from the exponent.

In mathematical terms, if you have a function \(f(x) = x^n\), its derivative \(f'(x)\) is \(nx^{n-1}\). For example, given the function \(c = 2s^{\frac{1}{2}} - 1\), we apply the rule to the \(s^{\frac{1}{2}}\) term.

• Multiply by \(\frac{1}{2}\) to get \(2 \times \frac{1}{2} = 1\).
• Reduce the exponent by one to obtain \(s^{-\frac{1}{2}}\).

Thus, the derivative of \(2s^{\frac{1}{2}}\) using the power rule is \(s^{-\frac{1}{2}}\). This simplification is crucial in subsequent steps, such as finding the actual value of the derivative at specific points.

Square Root Function

Square root functions are quite common in mathematics, and you can recognize them by the radical symbol \(\sqrt{}\). In calculus, it is helpful to express these functions in exponent form to facilitate differentiation.The conversion from a square root to a power is represented by \(\sqrt{s} = s^{\frac{1}{2}}\). This representation is particularly useful because it allows us to apply the power rule seamlessly.By expressing square root functions as powers:

• We simplify differentiation.
• We make computation with other rules like the chain rule easier.

For instance, in our exercise, transforming \(\sqrt{s}\) to \(s^{\frac{1}{2}}\) was the key step that allowed the application of the power rule. Without changing this expression, calculating the derivative would be more complex. Remember that understanding how to manipulate and rewrite expressions is a powerful skill in calculus.

Evaluating Derivatives

Once you have found the derivative of a function, the next step often involves evaluating it at a specific point. This means substituting a given value into the derivative to find the slope or rate of change of the original function at that particular point.To evaluate the derivative of \(f(s) = s^{\frac{-1}{2}}\) at \(s = 25\):

• Substitute 25 into the derivative: \(s^{-\frac{1}{2}} = 25^{-\frac{1}{2}}\).
• This results in \(25^{-\frac{1}{2}} = \frac{1}{\sqrt{25}} = \frac{1}{5}\).

Evaluating derivatives in this way allows us to understand the behavior of functions at particular points. For example, in the context of a physical problem, this could indicate how quickly a quantity is changing at that point. This skill is beneficial in many scientific fields, where predicting and understanding changes are essential.