A) \[\frac{1+\tan x-x.{{\sec }^{2}}x}{{{(1+\tan x)}^{2}}}\]
B) \[\frac{1+\tan x+x.{{\sec }^{2}}x}{{{(1+\tan x)}^{2}}}\]
C) \[\frac{1-\tan x-x.{{\sec }^{2}}x}{{{(1+\tan x)}^{2}}}\]
D) None of these
Correct Answer: A
Solution :
[a] \[\because y=\frac{x}{(1+\tan x)}\] \[\therefore \frac{dy}{dx}=\frac{(1+\tan x).1-x(0+{{\sec }^{2}}x)}{{{(1+\tan x)}^{2}}}\] \[=\frac{1+tanx-x.se{{c}^{2}}x}{{{\left( 1+tanx \right)}^{2}}}\]
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