Question

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

• A. $\frac{a}{b}$
• B. $\frac{b}{a}$
• C. 1
• D. 0

Answer 1

The correct option is D 0

Solution:

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

Taking logarithm on both sides, we get

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

Differentiating on both sides, we get

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

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

or, $\frac{dy}{dx}=\frac{y}{x}.\frac{bx+ay}{bx+ay}$

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

Hence, D is the correct option.