Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find the second order derivatives of the functions given in Exercises 1 to 10. e ^ 6x cos 3x

Q7  Find the second order derivatives of the functions given in Exercises 1 to 10. 

       e ^{6x}\cos 3x

 

Answers (1)

best_answer

Given function is
y= e^{6x}\cos 3x
Now, differentiation w.r.t. x
\frac{dy}{dx}=6e^{6x}.\cos 3x +e^{6x}.(-3\sin 3x)= e^{6x}(6\cos 3x-3\sin 3x)
Now, second order derivative is
\frac{d^2y}{dx^2}= 6e^{6x}(6\cos3x-3\sin3x)+e^{6x}(6.(-3\sin3x)-3.3\cos3x)
          = 6e^{6x}(6\cos3x-3\sin3x)-e^{6x}(18\sin3x+9\cos3x)
          e^{6x}(27\cos3x-36\sin3x) = 9e^{6x}(3\cos3x-4\sin3x)
Therefore,  second order derivative is \frac{dy}{dx} = 9e^{6x}(3\cos3x-4\sin3x)

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

CBSE introduces CCTV policy in board exams 2025; schools fixed as centres must have facility
anam-khan

CBSE gives parents, schools till October 16 to register students in Classes 9, 11
anam-khan

CBSE Board Exam 2025: Class 9, 11 registration begins today; fee details, guidelines

Latest Question