Calculus - Differentiation

Differentiation

Shubhodip ·Nov 8 '11 at 23:00

x^3 + x^2 - 1>x^3

x^2 - 1>0
x^2 >1

is that wat u mean ?

Edit: @man111:Yeah .. its alright..i had a typo ..which i have corrected..

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1708

man111 singh ·Nov 8 '11 at 23:02

$\hspace{-16}$Let $\mathbf{u=\frac{1}{4}.x^4+\frac{1}{3}.x^3-x}$ and $\mathbf{v=\frac{x^4}{4}}$\\\\ Now Given that $\mathbf{\frac{du}{dx}<\frac{dv}{dx}}$\\\\ $\mathbf{x^3+x^2-1<x^3}$\\\\ $\mathbf{x^2-1<0}$\\\\ $\mathbf{(x-1).(x+1)<0}$\\\\ So $\mathbf{x\in (-\infty\;,-1)\cup (+1\;,\infty)}$$

Shubhodip i have not seen your solution.

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Abhinav Gupta ·Nov 9 '11 at 20:09

Thanks guys . Got my mistake.I was not differentiating R.H.S..

By the way ,MAN111 how do you type these solutions? I mean which software do you use?

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