Differentiation Question 256
Question: If $ y=\log x.{e^{(\tan x+x^{2})}}, $ then $ \frac{dy}{dx}= $
[AI CBSE 1985]
Options:
A) $ {e^{(\tan x+x^{2})}}[ \frac{1}{x}+({{\sec }^{2}}x+x)\log x ] $
B) $ {e^{(\tan x+x^{2})}}[ \frac{1}{x}+({{\sec }^{2}}x-x)\log x ] $
C) $ {e^{(\tan x+x^{2})}}[ \frac{1}{x}+({{\sec }^{2}}x+2x)\log x ] $
D) $ {e^{(\tan x+x^{2})}}[ \frac{1}{x}+({{\sec }^{2}}x-2x)\log x ] $
Show Answer
Answer:
Correct Answer: C
Solution:
$ y=\log x.{e^{(\tan x+{x^{2)}}}} $
$ \therefore \frac{dy}{dx}={e^{(\tan x+x^{2})}}.\frac{1}{x}+\log x.{e^{(\tan x+x^{2})}}({{\sec }^{2}}x+2x) $
$ ={e^{(\tan x+x^{2})}}[ \frac{1}{x}+({{\sec }^{2}}x+2x)\log x ] $ .
BETA
