Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Differentiate the following w.r.t. x: e ^ x ^1 / 2

7. Differentiate the following w.r.t. x: 

\sqrt { e ^{ \sqrt x }} , x > 0

Answers (1)

best_answer

Given function is
f(x)=\sqrt { e ^{ \sqrt x }}
Lets take g(x ) = \sqrt x
Now, our function reduces to 
f(x) = \sqrt {e^{g(x)}}
Now,
f^{'}(x) = g^{'}(x).\frac{1}{2\sqrt{e^{g(x)}}}.\frac{d({e^{g(x)}})}{dx} = g{'}(x).\frac{1}{2\sqrt{e^{g(x)}}}.{e^{g(x)}} = \frac{g^{'}(x).e^{g(x)}}{2.\sqrt{e^{g(x)}}} = \frac{g^{'}(x).e^{\sqrt x}}{2.\sqrt{e^{\sqrt x}}}                   -(i)
And
g(x)=\sqrt x\\\Rightarrow g^{'}(x) = \frac{(\sqrt x)}{dx}=\frac{1}{2\sqrt x} 
Put this value in our equation (i)
f^{'}(x) =\frac{e^{\sqrt x}}{2\sqrt x.2.\sqrt{e^{\sqrt x}}} = \frac{e^{\sqrt x}}{4\sqrt{xe^{\sqrt x}}}
Therefore, the answer is \frac{e^{\sqrt x}}{4\sqrt{xe^{\sqrt x}}}.\ \ x>0

Posted by

Gautam harsolia

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question