A) \[\frac{a}{b}\]
B) \[\frac{b}{a}\]
C) 1
D) 0
Correct Answer: D
Solution :
| [d] \[{{x}^{a}}{{y}^{b}}={{(x-y)}^{a+b}}\] |
| Taking log both the sides |
| \[\log \left( {{x}^{a}}{{y}^{b}} \right)=\log {{(x-y)}^{(a+b)}}\] |
| \[a\log x+b\log \,y=(a+b)\log (x-y)\] |
| differentiating both sides w.r.t ?x?. |
| \[\frac{a}{x}+\frac{b}{y}\frac{dy}{dx}=\frac{(a+b)}{(x-y)}\left[ 1-\frac{dy}{dx} \right]\] |
| \[\frac{dy}{dx}\left[ \frac{b}{y}+\frac{a+b}{x-y} \right]=\frac{a+b}{x-y}-\frac{a}{x}\] |
| \[\frac{dy}{dx}\left[ \frac{bx-by+ay+by}{y(x-y)} \right]=\frac{ax+bx-ax+ay}{x(x-y)}\] |
| \[\frac{dy}{dx}\left[ \frac{bx+ay}{y} \right]=\frac{bx+ay}{x}\] |
| \[\frac{dy}{dx}=\frac{y}{x};\frac{dy}{dx}-\frac{y}{x}=0\] |
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