Direction: For the next two (2) items that follow: Consider the equation x + |y| = 2y
What is the derivative of y as a function of x with respect to x for x < 0?
Finding the Derivative with Absolute Value
The problem asks us to find the derivative of \(y\) with respect to \(x\), denoted as \(\frac{dy}{dx}\), for the equation \(x + |y| = 2y\), under the specific condition that \(x < 0\).
Analyzing the Equation \(x + |y| = 2y\)
The equation involves an absolute value, \(|y|\). The definition of absolute value is:
- \(|y| = y\) if \(y \geq 0\)
- \(|y| = -y\) if \(y < 0\)
We need to determine which case applies given the condition \(x < 0\).
Case 1: Assume \(y \geq 0\)
If \(y \geq 0\), the equation becomes:
\(x + y = 2y\)
Subtracting \(y\) from both sides gives:
\(x = 2y - y\)
\(x = y\)
If \(x = y\) and we are given \(x < 0\), this implies \(y < 0\). However, this contradicts our initial assumption that \(y \geq 0\). Therefore, the case \(y \geq 0\) is not possible when \(x < 0\).
Case 2: Assume \(y < 0\)
If \(y < 0\), the equation becomes:
\(x + (-y) = 2y\)
\(x - y = 2y\)
Adding \(y\) to both sides gives:
\(x = 2y + y\)
\(x = 3y\)
If \(x = 3y\) and we are given \(x < 0\), this implies \(3y < 0\), which means \(y < 0\). This is consistent with our initial assumption that \(y < 0\). Therefore, for \(x < 0\), the relationship between \(x\) and \(y\) is described by the equation \(x = 3y\).
Finding the Derivative \(\frac{dy}{dx}\)
For \(x < 0\), the equation simplifying the relationship between \(x\) and \(y\) is \(x = 3y\). We need to find the derivative of \(y\) with respect to \(x\).
We can easily express \(y\) as a function of \(x\) from this equation:
\(y = \frac{1}{3}x\)
Now, we can differentiate \(y\) with respect to \(x\):
\(\frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{3}x\right)\)
Using the constant multiple rule and the power rule for differentiation:
\(\frac{dy}{dx} = \frac{1}{3} \frac{d}{dx}(x)\)
\(\frac{dy}{dx} = \frac{1}{3} \times 1\)
\(\frac{dy}{dx} = \frac{1}{3}\)
Alternatively, we could use implicit differentiation directly on the equation \(x = 3y\). Differentiating both sides with respect to \(x\):
\(\frac{d}{dx}(x) = \frac{d}{dx}(3y)\)
\(1 = 3 \frac{dy}{dx}\)
Solving for \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = \frac{1}{3}\)
Both methods yield the same result.
Summary of Steps for Finding the Derivative
- Identify the equation and the condition on \(x\).
- Analyze the absolute value term based on possible signs of \(y\).
- Use the given condition (\(x < 0\)) to determine which case for \(|y|\) is valid.
- Rewrite the original equation without the absolute value for the valid case.
- Differentiate the simplified equation with respect to \(x\) to find \(\frac{dy}{dx}\).
| Condition on \(x\) | Valid Case for \(y\) | Simplified Equation | Derivative \(\frac{dy}{dx}\) |
|---|---|---|---|
| \(x < 0\) | \(y < 0\) | \(x = 3y\) | \(\frac{1}{3}\) |
Therefore, for \(x < 0\), the derivative of \(y\) as a function of \(x\) with respect to \(x\) is \(\frac{1}{3}\).
Revision Table - Derivative Calculation
| Concept | Description | Application in this Problem |
|---|---|---|
| Absolute Value | Defines output based on input sign. \(|a|=a\) if \(a \ge 0\), \(|a|=-a\) if \(a < 0\). | Used to split the equation based on the sign of \(y\). |
| Function Domain/Conditions | The set of inputs for which a function or equation is defined or considered. | The condition \(x < 0\) helped determine the valid case for the absolute value \(|y|\). |
| Differentiation | Finding the rate of change of a function. \(\frac{dy}{dx}\) is the derivative of \(y\) with respect to \(x\). | Applied to the simplified equation \(y = \frac{1}{3}x\) (or \(x=3y\)) to find \(\frac{dy}{dx}\). |
| Implicit Differentiation | A technique to differentiate equations relating two or more variables without explicitly solving for one variable in terms of the other. | Could be used on \(x = 3y\) by differentiating both sides with respect to \(x\). |
Additional Information - Absolute Value Equations and Differentiation
Equations involving absolute values often require careful analysis of cases. When differentiating such equations implicitly or explicitly, it is crucial to first simplify the equation by considering the sign of the term inside the absolute value.
For the equation \(x + |y| = 2y\), the relationship between \(x\) and \(y\) depends on whether \(y\) is positive or negative:
- If \(y \ge 0\), then \(x = y\). This means the graph of \(y\) vs \(x\) is a line with slope 1.
- If \(y < 0\), then \(x = 3y\). This means the graph of \(y\) vs \(x\) is a line with slope \(\frac{1}{3}\).
The original equation actually describes two different lines in the \(xy\)-plane, separated at \(y=0\) (and thus \(x=0\)). The problem specifically asks about the derivative for \(x < 0\). Based on our analysis, when \(x < 0\), \(y\) must also be less than 0. Therefore, we are dealing with the part of the graph defined by \(y = \frac{1}{3}x\).
The derivative \(\frac{dy}{dx}\) represents the slope of the tangent line to the curve at a given point. For the linear relationship \(y = \frac{1}{3}x\) (which holds for \(x < 0\)), the slope is constant and equal to \(\frac{1}{3}\) for all \(x\) in that domain.
Similar Questions
If \({\rm{y}} = {\cos ^{ - 1}}\left( {\frac{{2{\rm{x}}}}{{1 + {{\rm{x}}^2}}}} \right)\) , then \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\) is equal to
What is the derivative of log 10 (5x 2+ 3) with respect to x?
The derivative of In(x + sin x) with respect to (x + cos x) is
If y = \(\frac{x \sqrt{x^2−16}}{2} − 8 \ln\left|x + \sqrt{x^2−16}\right|\) , then what is \(\frac{\text{dy}}{\text{dx}}\) equal to ?
If e θϕ = c + 4θϕ, where c is an arbitrary constant and ϕ is a function of θ, then what is ϕ dθ equal to?
If \(\rm x^m y^n =a^{m+n}\) , then what is \(\rm \dfrac{dy}{dx}\) equal to?
What is the derivative of sec 2(tan -1 x) with respect to x?
A function is defined in (0, ∞) by \( f(x) = \begin{cases} 1-x^2 & for& , 0 < x \leq 1 \quad \\ In \ x & for &, 1 < x \leq 2 \\ In \ 2 - 1 + 0.5x & for &, 2 < x < \infty \end{cases} \)
Which one of the following is correct in respect of the derivative of the function, i.e. f’(x)?Let f(x + y) = f(x) f(y) for all x and y. Then what is f’(5) equal to [where f’(x) is the derivative of f(x)]?
Which of the following statements are not correct?
1. y as a function of x is not defined for all real x.
2. y as a function of x is not continuous at x = 0
3. y as a function of x is differentiable for all x.
Select the correct answer using the code given below
Important Questions from Evaluation of derivatives
Differentiate {-log (log x), x > 1} with respect to x
Find the derivation of f(x) = 1/x2
- The derivative of the function f(x) = -3x2 + 6x - 4 is given by:
Differential coefficient of log10 x with respect to logx 10 is
The derivatives of (x3 + ex + 3x + cot x) with respect to x is
