Derivative of Square Root of 1-2x Explained

The derivative of the function \( f(x) = \sqrt{1-2x} \) can be calculated using the chain rule. The steps are detailed below:

Step-by-Step Calculation

1. Rewrite the square root function in exponential form:

   \[ f(x) = \sqrt{1-2x} = (1-2x)^{1/2} \]

2. Apply the chain rule:

   \[ \frac{d}{dx} \left[ (1-2x)^{1/2} \right] = \frac{1}{2}(1-2x)^{-1/2} \cdot \frac{d}{dx}(1-2x) \]

3. Differentiate the inner function:

   \[ \frac{d}{dx}(1-2x) = -2 \]

4. Combine the results:

   \[ \frac{d}{dx} \left[ (1-2x)^{1/2} \right] = \frac{1}{2}(1-2x)^{-1/2} \cdot (-2) \]

   \[ = -\frac{1}{(1-2x)^{1/2}} \]

   \[ = -\frac{1}{\sqrt{1-2x}} \]

Summary

The derivative of \( f(x) = \sqrt{1-2x} \) is:

\[ f'(x) = -\frac{1}{\sqrt{1-2x}} \]

Examples and Applications

• This derivative can be used to find the slope of the tangent line to the curve \( y = \sqrt{1-2x} \) at any point \( x \).
• Understanding this derivative is essential for solving problems in physics and engineering where square root functions appear.

Further Reading

For more examples and detailed explanations, visit and .

The derivative of the function \( \sqrt{1 - 2x} \) is a fundamental concept in calculus, often encountered in differentiation problems involving the chain rule. Understanding how to derive this function step by step can deepen your comprehension of differentiation techniques. This section will guide you through the process of finding the derivative of \( \sqrt{1 - 2x} \) using the chain rule and power rule, which are essential tools in calculus.

Let's start with the function:

\[ y = \sqrt{1 - 2x} \]

To differentiate \( y \), we first rewrite the square root function in exponent form:

\[ y = (1 - 2x)^{1/2} \]

Next, we apply the chain rule. The chain rule states that if you have a composite function, the derivative is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.

Let:

\[ u = 1 - 2x \]

Then,

\[ y = u^{1/2} \]

Now, differentiate the outer function with respect to \( u \):

\[ \frac{dy}{du} = \frac{1}{2} u^{-1/2} \]

Next, differentiate the inner function with respect to \( x \):

\[ \frac{du}{dx} = -2 \]

Combine these results using the chain rule:

\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]

\[ \frac{dy}{dx} = \frac{1}{2} (1 - 2x)^{-1/2} \cdot (-2) \]

Simplify the expression:

\[ \frac{dy}{dx} = \frac{-1}{(1 - 2x)^{1/2}} \]

Or equivalently,

\[ \frac{dy}{dx} = \frac{-1}{\sqrt{1 - 2x}} \]

Thus, the derivative of \( \sqrt{1 - 2x} \) is:

\[ \boxed{\frac{-1}{\sqrt{1 - 2x}}} \]

To find the derivative of \( \sqrt{1-2x} \), we can use the chain rule. This method allows us to differentiate composite functions by differentiating the outer function and then the inner function.

1. Rewrite the function using fractional exponents:

   \[ y = \sqrt{1-2x} = (1-2x)^{1/2} \]

2. Identify the outer and inner functions:
   • Outer function: \( f(u) = u^{1/2} \)
   • Inner function: \( u = 1-2x \)
3. Differentiate the outer function with respect to \( u \):

   \[ \frac{d}{du}(u^{1/2}) = \frac{1}{2}u^{-1/2} = \frac{1}{2\sqrt{u}} \]

4. Differentiate the inner function with respect to \( x \):

   \[ \frac{d}{dx}(1-2x) = -2 \]

5. Apply the chain rule:

   \[ \frac{dy}{dx} = \frac{d}{du}(u^{1/2}) \cdot \frac{du}{dx} \]

   Substitute \( u = 1-2x \):

   \[ \frac{dy}{dx} = \frac{1}{2\sqrt{1-2x}} \cdot (-2) \]

6. Simplify the expression:

   \[ \frac{dy}{dx} = -\frac{1}{\sqrt{1-2x}} \]

Therefore, the derivative of \( \sqrt{1-2x} \) is \( -\frac{1}{\sqrt{1-2x}} \).

To better understand how to find the derivative of the square root of \(1-2x\), we will go through detailed, step-by-step examples. The process will involve using the chain rule and simplifying the result. Let's dive into the examples:

• Example 1: Find the derivative of \( \sqrt{1-2x} \).

  1. Let \( f(x) = \sqrt{1-2x} \). This can be rewritten as \( f(x) = (1-2x)^{1/2} \).
  2. Use the chain rule: \( \frac{d}{dx}[g(h(x))] = g'(h(x)) \cdot h'(x) \), where \( g(u) = u^{1/2} \) and \( h(x) = 1-2x \).
  3. Differentiate \( g(u) = u^{1/2} \) to get \( g'(u) = \frac{1}{2} u^{-1/2} = \frac{1}{2\sqrt{u}} \).
  4. Differentiate \( h(x) = 1-2x \) to get \( h'(x) = -2 \).
  5. Combine the results: \( f'(x) = g'(h(x)) \cdot h'(x) = \frac{1}{2\sqrt{1-2x}} \cdot (-2) = \frac{-1}{\sqrt{1-2x}} \).

• Example 2: Find the second derivative of \( \sqrt{1-2x} \).

  1. We have already found the first derivative: \( f'(x) = \frac{-1}{\sqrt{1-2x}} \).
  2. Rewrite the first derivative: \( f'(x) = -(1-2x)^{-1/2} \).
  3. Use the chain rule again to differentiate: \( f''(x) = \frac{d}{dx} [-(1-2x)^{-1/2}] \).
  4. Differentiate \( -(1-2x)^{-1/2} \): \( f''(x) = -\frac{1}{2} \cdot (1-2x)^{-3/2} \cdot (-2) = \frac{1}{(1-2x)^{3/2}} \).

• Example 3: Verify the derivative of \( \sqrt{1-2x} \) using implicit differentiation.

  1. Let \( y = \sqrt{1-2x} \).
  2. Square both sides: \( y^2 = 1-2x \).
  3. Differentiating both sides with respect to \( x \): \( 2y \cdot \frac{dy}{dx} = -2 \).
  4. Solve for \( \frac{dy}{dx} \): \( \frac{dy}{dx} = \frac{-2}{2y} = \frac{-1}{y} \).
  5. Substitute back \( y = \sqrt{1-2x} \): \( \frac{dy}{dx} = \frac{-1}{\sqrt{1-2x}} \).

The key concepts involved in finding the derivative of the square root of 1-2x include understanding the power rule, the chain rule, and the simplification process. Below is a detailed explanation of these concepts with step-by-step examples.

Power Rule

The power rule states that if you have a function of the form \( f(x) = x^n \), the derivative \( f'(x) \) is given by:

\[
\frac{d}{dx} x^n = n x^{n-1}
\]

For example, if \( f(x) = x^{1/2} \), then:

\[
\frac{d}{dx} x^{1/2} = \frac{1}{2} x^{-1/2} = \frac{1}{2 \sqrt{x}}
\]

Chain Rule

The chain rule is used when differentiating composite functions. If you have a function \( g(x) = f(u(x)) \), where \( u(x) \) is another function, the chain rule states:

\[
\frac{d}{dx} f(u(x)) = f'(u(x)) \cdot u'(x)
\]

For the function \( \sqrt{1-2x} \), let \( u(x) = 1-2x \). Then \( \sqrt{1-2x} \) becomes \( (1-2x)^{1/2} \). The chain rule requires us to differentiate the outer function and then multiply by the derivative of the inner function:

\[
\frac{d}{dx} (1-2x)^{1/2} = \frac{1}{2} (1-2x)^{-1/2} \cdot (-2)
\]

Simplification

After applying the chain rule, the next step is to simplify the resulting expression. Continuing from the previous step:

\[
\frac{1}{2} (1-2x)^{-1/2} \cdot (-2) = -\frac{1}{\sqrt{1-2x}}
\]

This is the simplified form of the derivative of \( \sqrt{1-2x} \).

Summary

• Apply the power rule: Rewrite the square root function as a power.
• Use the chain rule: Differentiate the outer function and multiply by the derivative of the inner function.
• Simplify the expression: Combine like terms and simplify the result to obtain the final derivative.

To deepen your understanding of the derivative of square root functions, particularly √(1-2x), exploring related mathematical concepts can be immensely helpful. Here are some topics that are closely related and essential for a comprehensive grasp of derivatives involving square roots:

• Integration of Root Functions:

  Understanding the integration of functions involving square roots is a natural progression from differentiation. For example, integrating functions like √(1-2x) or similar expressions often requires substitution techniques and familiarity with integral tables.

• Derivative of Composite Functions:

  Composite functions involve applying the chain rule extensively. For instance, finding the derivative of functions such as f(g(x)) where g(x) = 1-2x and f(x) = √x requires a step-by-step approach using the chain rule.

• Graphing Derivatives:

  Visualizing derivatives through graphing helps in understanding the behavior of the function's rate of change. Plotting the derivative of √(1-2x) shows how the slope of the tangent line varies along the curve.

• Applications in Physics and Engineering:

  Derivatives of square root functions are frequently used in physics and engineering. For example, they appear in formulas for motion under gravity, electrical circuits, and material stress analysis.

• Higher-Order Derivatives:

  Exploring the second and higher-order derivatives of square root functions can provide deeper insights into the concavity and inflection points of the functions, which are crucial for advanced calculus and analysis.

• Inverse Functions and Their Derivatives:

  Understanding the derivatives of inverse functions is another critical area. For example, the inverse of √(1-2x) and its differentiation involve interesting applications of the chain rule and implicit differentiation.

• What is the general formula for the derivative of a square root function?

  To find the derivative of a square root function \( \sqrt{f(x)} \), use the chain rule. The general formula is:
  \[
  \frac{d}{dx} \left( \sqrt{f(x)} \right) = \frac{f'(x)}{2\sqrt{f(x)}}
  \]
  For example, if \( f(x) = 1 - 2x \), then \( \frac{d}{dx} \left( \sqrt{1 - 2x} \right) = \frac{-2}{2\sqrt{1 - 2x}} = \frac{-1}{\sqrt{1 - 2x}} \).

• How can the derivative of root functions be applied in real-world scenarios?

  Derivatives of root functions are used in various fields such as physics, engineering, and economics. For instance, in physics, the square root function can describe the relationship between kinetic energy and velocity. In economics, it can model cost functions where costs increase at a decreasing rate.

• What are the common mistakes to avoid when differentiating square root functions?

  Common mistakes include:
  

  
  
  • Forgetting to apply the chain rule correctly, especially when the inner function is more complex.
  
  
  • Incorrectly simplifying the derivative, such as misapplying the power rule.
  
  
  • Neglecting to multiply by the derivative of the inner function \( f(x) \).
  
  
  • Making algebraic errors when combining terms.