logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
+1 vote

Differentiate w.r.t. \(x\) the function in \( x^{\large x^2-3} + ( x - 3 )^{\large x^2} \), for \(x > 3\)

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $[(uv)'=u'v+uv']$
Step 1:
It is of the form $u+v$
$\Rightarrow u=x^{\large x^2-3}$
$v=(x-3)^{\large x^2}$
Now $u=x^{\large x^2-3}$
Taking $\log$ on both sides
$\log u=\log x^{\large x^2-3}=(x^2-3)\log x$
Step 2:
Differentiating with respect to $x$
$\large\frac{1}{u}\frac{du}{dx}$$=2x\log x+(x^2-3)\large\frac{1}{x}$
$\large\frac{du}{dx}$$=u\big(2x\log x+\large\frac{x^2-3}{x}\big)$
$\quad\;=x^{\large x^2-3}[2x\log x+\large\frac{x^2-3}{x}]$
Step 3:
Now $v=(x-3)^{\Large x^{\Large 2}}$
Taking $\log$ on both sides
Now $\log v=\log(x-3)^{\Large x^{\Large 2}}$
Differentiating both sides
$\large\frac{1}{v}\frac{dv}{dx}=$$[2x\log(x-3)+\large\frac{x^2}{x-3}]$
$\large\frac{dv}{dx}$$=v.[2x\log(x-3)+\large\frac{x^2}{x-3}]$
$\large\frac{dv}{dx}$$=(x-3)^{\Large x^{\Large 2}}.[2x\log(x-3)+\large\frac{x^2}{x-3}]$
Step 4:
$\large\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
$\quad\;=x^{\Large x^{\Large 2}- 3}[$$2x\log x+\large\frac{x^2-3}{x}]$$+(x-3)^{\Large x^{\Large 2}}.[2x\log(x-3)+\large\frac{x^2}{x-3}]$
answered May 14, 2013 by sreemathi.v
 

Related questions

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
...