A) \[\frac{1-x}{x}\]
B) \[\frac{1}{x}\]
C) \[\frac{x}{1+x}\]
D) \[\frac{1+x}{x}\]
Correct Answer: A
Solution :
Given, \[x={{e}^{y+x}}\] \[\Rightarrow \] \[log\text{ }x=x+y\] On differentiating w.r.t.\[x,\]we get \[\frac{1}{x}=1+\frac{dy}{dx}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{1-x}{x}\]You need to login to perform this action.
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