Differentiation Question 410
Question: If $ x^{2}e^{y}+2xye^{x}+13=0 $ , then dy/dx =
[RPET 1987]
Options:
A) $ \frac{2x{e^{y-x}}+2y(x+1)}{x(x{e^{y-x}}+2)} $
B) $ \frac{2x{e^{x-y}}+2y(x+1)}{x(x{e^{y-x}}+2)} $
C) $ -\frac{2x{e^{y-x}}+2y(x+1)}{x(x{e^{y-x}}+2)} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{dy}{dx}=-\frac{\partial f/\partial x}{\partial f/\partial y}=-\frac{2xe^{y}+2xye^{x}+2ye^{x}}{x^{2}e^{y}+2xe^{x}} $
$ =-\frac{2x{e^{y-x}}+2y(x+1)}{x(x{e^{y-x}}+2)} $ , [Dividing $ N^{r} $ and $ D^{r} $ by $ e^{x} $ ].
BETA
