Differentiation Question 113
Question: If $ y\sec x+\tan x+x^{2}y=0 $ , then $ \frac{dy}{dx} $ =
[DSSE 1981; CBSE 1981]
Options:
A) $ \frac{2xy+{{\sec }^{2}}x+y\sec x\tan x}{x^{2}+\sec x} $
B) $ -\frac{2xy+{{\sec }^{2}}x+\sec x\tan x}{x^{2}+\sec x} $
C) $ -\frac{2xy+{{\sec }^{2}}x+y\sec x\tan x}{x^{2}+\sec x} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ y\sec x+\tan x+x^{2}y=0 $
$ \Rightarrow \sec x\frac{dy}{dx}+y\sec x\tan x+{{\sec }^{2}}x+2xy+x^{2}\frac{dy}{dx}=0 $
Therefore $ \frac{dy}{dx}=-\frac{2xy+{{\sec }^{2}}x+y\sec x\tan x}{x^{2}+\sec x} $ .
BETA
