logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

Find the second order derivatives of the function $(e^{\large x} \sin 5x)$

$\begin{array}{1 1} e^{\large x}[24\sin 5x-10\cos 5x] \\ e^{\large x}(5\cos 5x-\sin 5x) \\ e^{\large x}[-24\sin 5x+10\cos 5x] \\e^{\large x}(5\cos 5x+\sin 5x)\end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $y=f(x)$
  • $\large\frac{dy}{dx}$$=f'(x)$
  • $\large\frac{d^2y}{dx^2}=\frac{d}{dx}\big(\frac{dy}{dx}\big)$
  • $\large\frac{d}{dx}$$(e^{\large x})=e^{\large x}$
Step 1:
$y=e^{\large x}\sin 5x$
Differentiating with respect to $x$
$\large\frac{dy}{dx}$$=e^{\large x}\large\frac{d}{dx}$$\sin 5x+\sin 5x.e^{\large x}.$
$\quad\;=e^{\large x}\cos 5x.5+\sin 5x.e^{\large x}$
$\quad\;=e^{\large x}(5\cos 5x+\sin 5x)$
Step 2:
$\large\frac{d^2y}{dx^2}$$=e^{\large x}.\large\frac{d}{dx}$$(5\cos 5x+\sin 5x)+\large\frac{d}{dx}$$(e^{\large x})$$.5\cos 5x+\sin 5x$
$\quad\;=e^{\large x}[-5.\sin 5x.5+\cos 5x.5]+[5\cos 5x+\sin 5x].e^{\large x}$
$\quad\;=e^{\large x}[-25.\sin 5x+5\cos 5x]+e^{\large x}[5\cos 5x+\sin 5x]$
$\quad\;=e^{\large x}[-25\sin 5x+5\cos 5x+5\cos 5x+\sin 5x]$
$\quad\;=e^{\large x}[-24\sin 5x+10\cos 5x]$
answered May 10, 2013 by sreemathi.v
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...