If x ay b= (x - y) a+b , then the value of $\frac{{{\rm{dy}}}}{{{\rm{dx}}}} - \frac{{\rm{y}}}{{\rm{x}}}$ is equal to

Question

Prepp Team · Accepted Answer

Understanding the Implicit Differentiation Problem

We are given an implicit relationship between $x$ and $y$: $x^a y^b = (x - y)^{a+b}$. Our goal is to find the value of the expression $\frac{{{\rm{dy}}}}{{{\rm{dx}}}} - \frac{{\rm{y}}}{{\rm{x}}}$.

This type of problem often involves implicit differentiation. When variables are multiplied with powers, taking logarithms before differentiating can significantly simplify the process.

Applying Logarithms to Simplify the Equation

Let's start with the given equation:

$x^a y^b = (x - y)^{a+b}$

Taking the natural logarithm ($\ln$) on both sides:

$\ln(x^a y^b) = \ln((x - y)^{a+b})$

Using the logarithm properties ($\ln(MN) = \ln M + \ln N$ and $\ln(M^P) = P \ln M$):

$\ln(x^a) + \ln(y^b) = (a + b) \ln(x - y)$

$a \ln x + b \ln y = (a + b) \ln(x - y)$

This form is much easier to differentiate.

Performing Implicit Differentiation

Now, we differentiate both sides of the simplified equation with respect to $x$. Remember that $\frac{d}{dx}(\ln u) = \frac{1}{u} \frac{du}{dx}$ and $\frac{d}{dx}(y) = \frac{dy}{dx}$.

$\frac{d}{dx}(a \ln x + b \ln y) = \frac{d}{dx}((a + b) \ln(x - y))$

$a \frac{d}{dx}(\ln x) + b \frac{d}{dx}(\ln y) = (a + b) \frac{d}{dx}(\ln(x - y))$

$a \cdot \frac{1}{x} + b \cdot \frac{1}{y} \frac{dy}{dx} = (a + b) \cdot \frac{1}{x - y} \cdot \frac{d}{dx}(x - y)$

$\frac{a}{x} + \frac{b}{y} \frac{dy}{dx} = \frac{a + b}{x - y} \cdot (1 - \frac{dy}{dx})$

Now, distribute the term on the right side:

$\frac{a}{x} + \frac{b}{y} \frac{dy}{dx} = \frac{a + b}{x - y} - \frac{a + b}{x - y} \frac{dy}{dx}$

Solving for $\frac{dy}{dx}$

Next, we group the terms containing $\frac{dy}{dx}$ on one side and the other terms on the other side:

$\frac{b}{y} \frac{dy}{dx} + \frac{a + b}{x - y} \frac{dy}{dx} = \frac{a + b}{x - y} - \frac{a}{x}$

Factor out $\frac{dy}{dx}$ from the terms on the left side:

$\frac{dy}{dx} \left( \frac{b}{y} + \frac{a + b}{x - y} \right) = \frac{a + b}{x - y} - \frac{a}{x}$

Find a common denominator for the expressions within the parenthesis on the left and on the right side:

Left side parenthesis: $\frac{b(x - y) + y(a + b)}{y(x - y)} = \frac{bx - by + ay + by}{y(x - y)} = \frac{bx + ay}{y(x - y)}$

Right side: $\frac{x(a + b) - a(x - y)}{x(x - y)} = \frac{ax + bx - ax + ay}{x(x - y)} = \frac{bx + ay}{x(x - y)}$

Substitute these simplified expressions back into the equation:

$\frac{dy}{dx} \left( \frac{bx + ay}{y(x - y)} \right) = \frac{bx + ay}{x(x - y)}$

To isolate $\frac{dy}{dx}$, divide both sides by the term multiplying $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{\frac{bx + ay}{x(x - y)}}{\frac{bx + ay}{y(x - y)}}$

Assuming $bx + ay \neq 0$ and $x \neq y$, we can simplify this expression:

$\frac{dy}{dx} = \frac{bx + ay}{x(x - y)} \cdot \frac{y(x - y)}{bx + ay}$

$\frac{dy}{dx} = \frac{y}{x}$

Calculating the Final Expression

Now we need to find the value of $\frac{{{\rm{dy}}}}{{{\rm{dx}}}} - \frac{{\rm{y}}}{{\rm{x}}}$. We found that $\frac{dy}{dx} = \frac{y}{x}$.

Substitute this into the expression:

$\frac{{{\rm{dy}}}}{{{\rm{dx}}}} - \frac{{\rm{y}}}{{\rm{x}}} = \frac{y}{x} - \frac{y}{x}$

$\frac{{{\rm{dy}}}}{{{\rm{dx}}}} - \frac{{\rm{y}}}{{\rm{x}}} = 0$

The value of the expression is 0.

Step	Action	Result
1	Start with the given equation	$x^a y^b = (x - y)^{a+b}$
2	Take log on both sides	$a \ln x + b \ln y = (a + b) \ln(x - y)$
3	Differentiate implicitly w.r.t. x	$\frac{a}{x} + \frac{b}{y} \frac{dy}{dx} = \frac{a + b}{x - y} (1 - \frac{dy}{dx})$
4	Rearrange to solve for $\frac{dy}{dx}$	$\frac{dy}{dx} = \frac{y}{x}$
5	Substitute $\frac{dy}{dx}$ into the target expression	$\frac{y}{x} - \frac{y}{x}$
6	Calculate the final value	0

Conclusion

By applying logarithms and implicit differentiation to the given equation $x^a y^b = (x - y)^{a+b}$, we found that $\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{\rm{y}}}{{\rm{x}}}$. Substituting this result into the expression $\frac{{{\rm{dy}}}}{{{\rm{dx}}}} - \frac{{\rm{y}}}{{\rm{x}}}$ yields 0.

Revision Table: Implicit Differentiation Key Concepts

Concept	Description
Implicit Function	A function where y is not explicitly expressed in terms of x (e.g., $x^2 + y^2 = 1$).
Implicit Differentiation	A technique used to find the derivative of an implicit function by differentiating each term with respect to the independent variable (usually x), remembering the chain rule for terms involving the dependent variable (y).
Chain Rule	$\frac{d}{dx}(f(y)) = f'(y) \frac{dy}{dx}$. For example, $\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$.
Logarithm Differentiation	A method used for differentiating functions raised to a power or functions involving products/quotients by taking the logarithm of both sides before differentiating.
Logarithm Properties	$\ln(MN) = \ln M + \ln N$ $\ln(\frac{M}{N}) = \ln M - \ln N$ $\ln(M^P) = P \ln M$

Additional Information: Applications of Derivatives

Derivatives, like $\frac{dy}{dx}$, represent the instantaneous rate of change of one variable with respect to another. They have numerous applications in various fields:

In Physics: Calculating velocity (derivative of position), acceleration (derivative of velocity).
In Engineering: Analyzing rates of change, optimization problems, designing systems.
In Economics: Determining marginal cost, marginal revenue, optimizing profit.
In Biology: Modeling population growth rates.
In Curve Sketching: Finding slopes of tangent lines, locating critical points (local maxima/minima), determining concavity.

Implicit differentiation is particularly useful when dealing with equations that define relationships between variables that are not easily solved for one variable in terms of the other, such as equations of circles or other complex curves.

Prepp Team · Answer

$\frac{{\rm{a}}}{{\rm{b}}}$

Prepp Team · Answer

$\frac{{\rm{b}}}{{\rm{a}}}$

Prepp Team · Answer

1