Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Differentiate the functions with respect to x in sin x ^ 2 + 5

1. Differentiate the functions with respect to x in 

\sin (x^2 +5 )

Answers (1)

best_answer

Given function is
f(x)=\sin (x^2 +5 )
when we differentiate it w.r.t. x.
Lets take t = x^2+5 . then,
f(t) = \sin t
\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx}                                          (By  chain rule)
\frac{df(t)}{dt} = \frac{d(\sin t )}{dt} = \cos t = \cos (x^2+5)
\frac{dt}{dx} = \frac{d(x^2+5 )}{dx} = 2x
Now,
\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} = \cos (x^2+5).2x
Therefore, the answer is 2x \cos (x^2+5)

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

Board Exam Date Sheet 2025: CBSE, UPMSP, MPBSE, Class 10, 12 board exam dates out; BSEB, CGBSE schedules soon
anam-khan

CBSE Date Sheet 2025 (Out) Live: Class 10, 12 board exam time table at cbse.gov.in; syllabus, sample papers
anam-khan

CBSE Class 12 board exam date sheet 2025 out; exams from February 15 to April 4

Latest Question