# Thread: quotient rule differentiation check !!

1. ## quotient rule differentiation check !!

$

f(x) = \frac{sinx}{x^2+1}

$

differentiation is:

$
f '(x) = \frac{(x^2+1)(cosx)-(sinx)(2x)}{(x^2+1)^2}
$

$
f '(x) = \frac{(cosx)-(sinx)(2x)}{x^2+1}
$

is this correct? thank u

2. Originally Posted by jvignacio
$

f(x) = \frac{sinx}{x^2+1}

$

differentiation is:

$
f '(x) = \frac{(x^2+1)(cosx)-(sinx)(2x)}{(x^2+1)^2}
$

$
f '(x) = \frac{(cosx)-(sinx)(2x)}{x^2+1}
$

is this correct? thank u
It is not. Here's why:

$
f '(x) = \frac{(x^2+1)(\cos x)-(\sin x)(2x)}{(x^2+1)^2}
$

In order to cancel out the $x^2+1$ 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:

$f'(x)=\frac{x^2\cos x-2x\sin x+\cos x}{(x^2+1)^2}$

However, IF you break up the expression into two different fractions, observe what happens:

$f'(x)=\frac{\cos x}{x^2+1}-\frac{2x\sin x}{(x^2+1)^2}$

Either of these answers would be correct.

I hope this makes sense!

--Chris

3. Originally Posted by Chris L T521
It is not. Here's why:

$
f '(x) = \frac{(x^2+1)(\cos x)-(\sin x)(2x)}{(x^2+1)^2}
$

In order to cancel out the $x^2+1$ 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:

$f'(x)=\frac{x^2\cos x-2x\sin x+\cos x}{(x^2+1)^2}$

However, IF you break up the expression into two different fractions, observe what happens:

$f'(x)=\frac{\cos x}{x^2+1}-\frac{2x\sin x}{(x^2+1)^2}$

Either of these answers would be correct.

I hope this makes sense!

--Chris

what do u mean by it must appear in both terms in the numerator? do u mean one has to be + and the other - to cancel em out? i thought they cud be on the numerator and the denominator to get cancelled out.

4. Originally Posted by jvignacio
what do u mean by it must appear in both terms in the numerator? do u mean one has to be + and the other - to cancel em out? i thought they cud be on the numerator and the denominator to get cancelled out.
$f'(x)=\frac{x^2\cos x-2x\sin x+\cos x}{(x^2+1)^2}$

Let's group the cos together.

$f'(x) = \frac{(x^2\cos{x}+\cos{x}) - 2x\sin x}{(x^2+1)^2}$

Split it into two fractions now:

$f'(x) = \frac{(x^2\cos x+\cos x)}{(x^2+1)^2} - \frac{2x\sin x}{(x^2+1)^2}$

Take \cos x common factor:
$f'(x) = \frac{\cos x(x^2+1)}{(x^2+1)^2} - \frac{2x\sin x}{(x^2+1)^2}$

Cancel:
$f'(x) = \frac{\cos x}{x^2+1} - \frac{2x\sin x}{(x^2+1)^2}$

5. Originally Posted by jvignacio
what do u mean by it must appear in both terms in the numerator? do u mean one has to be + and the other - to cancel em out? i thought they cud be on the numerator and the denominator to get cancelled out.
What Chris is saying is saying, more or less, is that $\frac{AB - C}{A^2} \neq \frac{B - C}{A}$. 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.

6. Originally Posted by Chop Suey
$f'(x)=\frac{x^2\cos x-2x\sin x+\cos x}{(x^2+1)^2}$

Let's group the cos together.

$f'(x) = \frac{(x^2\cos{x}+\cos{x}) - 2x\sin x}{(x^2+1)^2}$

Split it into two fractions now:

$f'(x) = \frac{(x^2\cos x+\cos x)}{(x^2+1)^2} - \frac{2x\sin x}{(x^2+1)^2}$

Take \cos x common factor:
$f'(x) = \frac{\cos x(x^2+1)}{(x^2+1)^2} - \frac{2x\sin x}{(x^2+1)^2}$

Cancel:
$f'(x) = \frac{\cos x}{x^2+1} - \frac{2x\sin x}{(x^2+1)^2}$
how did u get from

$f '(x) = (x^2+1)(cos x)-(sin x)(2x)$

to
$f'(x)=x^2\cos x-2x\sin x+\cos x$

7. Originally Posted by jvignacio
how did u get from

$f '(x) = (x^2+1)(cos x)-(sin x)(2x)$

to
$f'(x)=x^2\cos x-2x\sin x+\cos x$
Basic algebra again.

$f '(x) = (x^2+1)(cos x)-(sin x)(2x)$

$=x^2 \cos x + 1 \cos x - 2x \sin x$

(multiplication is distributive: (a+b)c = ac + bc, so put a=x^2, b= 1, c=cos x in the above)

8. Originally Posted by jvignacio
how did u get from

$f '(x) = (x^2+1)(cos x)-(sin x)(2x)$

to
$f'(x)=x^2\cos x-2x\sin x+\cos x$
The revision you've been advised to undertake includes how to expand ....

9. Originally Posted by jvignacio
what do u mean by it must appear in both terms in the numerator? do u mean one has to be + and the other - to cancel em out? i thought they cud be on the numerator and the denominator to get cancelled out.
It has to be a common factor to be cancelled. If you got:
$\frac{AB+C}{A}$

then it can be split into:

$\frac{AB}{A} + \frac{C}{A}$

The A in this case can be cancelled and it will become:
$B + \frac{C}{A}$

Another case..if you got:
$\frac{AB + AC}{A}$

Then, you can pull A common factor:
$\frac{A(B+C)}{A}$

and then cancel to get:
$B + C$

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!

10. Originally Posted by Matt Westwood
Basic algebra again.

$f '(x) = (x^2+1)(cos x)-(sin x)(2x)$

$=x^2 \cos x + 1 \cos x - 2x \sin x$

(multiplication is distributive: (a+b)c = ac + bc, so put a=x^2, b= 1, c=cos x in the above)
oh yesss i knew that, sorry i just didnt know u cud do that with trigonometric functions.

11. Originally Posted by Chop Suey
It has to be a common factor to be cancelled. If you got:
$\frac{AB+C}{A}$

then it can be split into:

$\frac{AB}{A} + \frac{C}{A}$

The A in this case can be cancelled and it will become:
$B + \frac{C}{A}$

Another case..if you got:
$\frac{AB + AC}{A}$

Then, you can pull A common factor:
$\frac{A(B+C)}{A}$

and then cancel to get:
$B + C$

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!
yes yes i understand wat u did. thanks man