$\displaystyle
f(x) = \frac{sinx}{x^2+1}
$
differentiation is:
$\displaystyle
f '(x) = \frac{(x^2+1)(cosx)-(sinx)(2x)}{(x^2+1)^2}
$
$\displaystyle
f '(x) = \frac{(cosx)-(sinx)(2x)}{x^2+1}
$
is this correct? thank u
$\displaystyle
f(x) = \frac{sinx}{x^2+1}
$
differentiation is:
$\displaystyle
f '(x) = \frac{(x^2+1)(cosx)-(sinx)(2x)}{(x^2+1)^2}
$
$\displaystyle
f '(x) = \frac{(cosx)-(sinx)(2x)}{x^2+1}
$
is this correct? thank u
It is not. Here's why:
$\displaystyle
f '(x) = \frac{(x^2+1)(\cos x)-(\sin x)(2x)}{(x^2+1)^2}
$
In order to cancel out the $\displaystyle x^2+1$ term, it must appear in both terms in the numerator. It doesn't, so we can't cancel it out.
You will have to leave the solution in the form:
$\displaystyle f'(x)=\frac{x^2\cos x-2x\sin x+\cos x}{(x^2+1)^2}$
However, IF you break up the expression into two different fractions, observe what happens:
$\displaystyle f'(x)=\frac{\cos x}{x^2+1}-\frac{2x\sin x}{(x^2+1)^2}$
Either of these answers would be correct.
I hope this makes sense!
--Chris
$\displaystyle f'(x)=\frac{x^2\cos x-2x\sin x+\cos x}{(x^2+1)^2}$
Let's group the cos together.
$\displaystyle f'(x) = \frac{(x^2\cos{x}+\cos{x}) - 2x\sin x}{(x^2+1)^2}$
Split it into two fractions now:
$\displaystyle f'(x) = \frac{(x^2\cos x+\cos x)}{(x^2+1)^2} - \frac{2x\sin x}{(x^2+1)^2}$
Take \cos x common factor:
$\displaystyle f'(x) = \frac{\cos x(x^2+1)}{(x^2+1)^2} - \frac{2x\sin x}{(x^2+1)^2}$
Cancel:
$\displaystyle f'(x) = \frac{\cos x}{x^2+1} - \frac{2x\sin x}{(x^2+1)^2}$
What Chris is saying is saying, more or less, is that $\displaystyle \frac{AB - C}{A^2} \neq \frac{B - C}{A}$. Substitute simple values for A, B and C if you don't see why.
Misunderstanding of basic algebra such as this just should not happen at the level of studying differential calculus. You are strongly advised to go back and revise all of the basic algebra that is assumed pre-requisite knowledge.
It has to be a common factor to be cancelled. If you got:
$\displaystyle \frac{AB+C}{A}$
then it can be split into:
$\displaystyle \frac{AB}{A} + \frac{C}{A}$
The A in this case can be cancelled and it will become:
$\displaystyle B + \frac{C}{A}$
Another case..if you got:
$\displaystyle \frac{AB + AC}{A}$
Then, you can pull A common factor:
$\displaystyle \frac{A(B+C)}{A} $
and then cancel to get:
$\displaystyle B + C$
I listed it step by step, so if you don't understand what I did, then something's really wrong and you must do what mr f suggested.
EDIT: Oh, I'm late!