Chapter 23: Problem 15
Find the derivative of each function. Check some by calculator. $$y=\sqrt{1-2 x}$$
Short Answer
Expert verified
\( \frac{dy}{dx} = -1 \cdot (1 - 2x)^{-1/2} \) or \( \frac{dy}{dx} = -\frac{1}{\sqrt{1-2x}} \)
Step by step solution
01
Identify the Function
Identify the function that needs to be differentiated, which is given as \( y = \sqrt{1 - 2x} \).
02
Rewrite the Function
Rewrite the square root function in exponential form: \( y = (1 - 2x)^{1/2} \).
03
Apply the Chain Rule
Use the chain rule for differentiation. If \( y = u^{n} \), and \( u \) is a function of \( x \), then the derivative, \( \frac{dy}{dx} \), is \( ny^{n-1} \cdot \frac{du}{dx} \). In this case, \( u = 1 - 2x \) and \( n = \frac{1}{2} \).
04
Differentiate the Inside Function
Differentiate \( u = 1 - 2x \) with respect to \( x \) to obtain \( \frac{du}{dx} = -2 \).
05
Apply the Derivative
Compute the derivative of \( y \) with respect to \( x \) using the chain rule as follows: \( \frac{dy}{dx} = \frac{1}{2}(1 - 2x)^{-1/2} \cdot (-2) \).
06
Simplify the Derivative
Simplify the expression to obtain the final derivative: \( \frac{dy}{dx} = -\frac{2}{2} \cdot (1 - 2x)^{-1/2} = -1 \cdot (1 - 2x)^{-1/2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
Understanding the chain rule is crucial when it comes to differentiating composite functions, such as square root functions. In the context of derivatives, a composite function occurs when one function is nested inside another. Think of it as a set of functions combined into one, where you first apply the inner function and then the outer function.
To differentiate such combined functionality, we use the chain rule, which tells us that the derivative of a composite function is the derivative of the outer function multiplied by the derivative of the inner function. The formal expression of the chain rule for a composite function with components of the form \(y = u(x)^{n}\) is \(\frac{dy}{dx} = n\cdot u^{n-1} \cdot \frac{du}{dx}\), where \(\frac{du}{dx}\) is the derivative of \(u\) with respect to \(x\).
For example, if we have \(y = \sqrt{1 - 2x}\) and we set \(u = 1 - 2x\) with \(n = \frac{1}{2}\), applying the chain rule gives us the derivative \(\frac{dy}{dx}\), which involves computing \(\frac{du}{dx}\) and plugging it into the formula. This approach simplifies the differentiation process and reveals how functions are connected on a deeper level.
To differentiate such combined functionality, we use the chain rule, which tells us that the derivative of a composite function is the derivative of the outer function multiplied by the derivative of the inner function. The formal expression of the chain rule for a composite function with components of the form \(y = u(x)^{n}\) is \(\frac{dy}{dx} = n\cdot u^{n-1} \cdot \frac{du}{dx}\), where \(\frac{du}{dx}\) is the derivative of \(u\) with respect to \(x\).
For example, if we have \(y = \sqrt{1 - 2x}\) and we set \(u = 1 - 2x\) with \(n = \frac{1}{2}\), applying the chain rule gives us the derivative \(\frac{dy}{dx}\), which involves computing \(\frac{du}{dx}\) and plugging it into the formula. This approach simplifies the differentiation process and reveals how functions are connected on a deeper level.
Exponential Form of Square Roots
When facing a square root function in calculus, it is often beneficial to convert it into its exponential form, making the process of differentiation more straightforward. For any square root function \(\sqrt{x}\), we can rewrite it as \(x^{1/2}\). This transforms it into a power function, where the exponent represents the root.
For instance, the function \(y = \sqrt{1 - 2x}\) can be rewritten in the exponential form as \(y = (1 - 2x)^{1/2}\). This allows us to apply the power rule and chain rule of differentiation directly, bypassing the need for a separate square root differentiation technique and making the function more manageable for further operations. Moreover, this approach is not limited to square roots; it applies to any root, by simply expressing the root degree as a fractional exponent.
For instance, the function \(y = \sqrt{1 - 2x}\) can be rewritten in the exponential form as \(y = (1 - 2x)^{1/2}\). This allows us to apply the power rule and chain rule of differentiation directly, bypassing the need for a separate square root differentiation technique and making the function more manageable for further operations. Moreover, this approach is not limited to square roots; it applies to any root, by simply expressing the root degree as a fractional exponent.
Derivative Computation
Derivative computation is a fundamental aspect of calculus that involves finding the instantaneous rate of change or the slope of the tangent line to the curve at any given point. To compute a derivative, we may use various rules of differentiation, such as the power rule, product rule, quotient rule, and the aforementioned chain rule.
For our function \(y = \sqrt{1 - 2x}\), after rewriting it as \(y = (1 - 2x)^{1/2}\), we apply derivative computation starting with the chain rule. We find the derivative of the outer function \(y^{\prime} = \frac{1}{2}(1 - 2x)^{-1/2}\) and then multiply it by the derivative of the inner function \(u^{\prime} = -2\). Finally, we simplify to get the final derivative \(\frac{dy}{dx} = -1\cdot(1 - 2x)^{-1/2}\). This process is the essence of derivative computation, systematically breaking down functions and applying rules to find the slope of function curves at any given point.
For our function \(y = \sqrt{1 - 2x}\), after rewriting it as \(y = (1 - 2x)^{1/2}\), we apply derivative computation starting with the chain rule. We find the derivative of the outer function \(y^{\prime} = \frac{1}{2}(1 - 2x)^{-1/2}\) and then multiply it by the derivative of the inner function \(u^{\prime} = -2\). Finally, we simplify to get the final derivative \(\frac{dy}{dx} = -1\cdot(1 - 2x)^{-1/2}\). This process is the essence of derivative computation, systematically breaking down functions and applying rules to find the slope of function curves at any given point.
