logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Model Papers
0 votes

If \( y=(x)^x+(\cos x)^{2x} \: find \: \large\frac{dy}{dx}\)

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $\large\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
Step 1:
$y=x^x+(\cos x)^{2x}$
Let $y=y_1+y_2$
Where $y_1=x^x$
$y_2=(\cos x)^{2x}$
Consider $y_1=x^x$
Take $\log$ on both sides
$\log y_1=x\log x$
Now differentiating w.r.t $x$ we get,
$\large\frac{1}{y_1}\frac{dy}{dx}$$=x.\large\frac{1}{x}$$+\log x.1$
$\large\frac{dy}{dx}$$=y_1[1+\log x]$
$\quad=x^x[1+\log x]$-----(1)
Step 2:
Consider $y_2=(\cos x)^{2x}$
Take $\log$ on both sides
$\log y_2=2x.\log(\cos x)$
Differentiating w.r.t $x$ we get,
$\large\frac{1}{y_2}\frac{dy}{dx}$$=2x.\large\frac{1}{\cos x}$$(-\sin x)+\log (\cos x)$
$\therefore \large\frac{dy}{dx}$$=y_2[\large\frac{-2x\sin x}{\cos x}$$+2\log(\cos x)]$
$\qquad=(\cos x)^{2x}[2\log x(\cos x)-2x.\tan x]$-----(2)
Step 3:
Combining equ(1) and equ(2) we get,
$\large\frac{dy}{dx}$$=x^x[1+\log x]+(\cos x)^{2x}[2\log(\cos x)-2x\tan x]$
answered Sep 25, 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
...