# simple diff quesstion

• Feb 17th 2010, 03:38 AM
decoy808
simple diff quesstion
want to check a question and solution.

differentiate with respect to x

$\displaystyle y= (x^3+4)/x$

= $\displaystyle x^2+4x^-{^1}$
( this is the line im confused about why does the x^3 become x^2 and not 3x) or is it wrong.

$\displaystyle dy/dx = 2x-4x^{-2}$

many thanks
• Feb 17th 2010, 03:40 AM
Prove It
Quote:

Originally Posted by decoy808
want to check a question and solution.

differentiate with respect to x

$\displaystyle y= (x^3+4)/x$

= $\displaystyle x^2+4x^-{^1}$
( this is the line im confused about why does the x^3 become x^2 and not 3x) or is it wrong.

$\displaystyle dy/dx = 2x-4x^{-2}$

many thanks

Note that $\displaystyle \frac{x^3 + 4}{x} = \frac{x^3}{x} + \frac{4}{x} = x^2 + 4x^{-1}$.

So when you take the derivative, it will be $\displaystyle 2x - 4x^{-2} = 2x - \frac{4}{x^2}$.
• Feb 17th 2010, 03:42 AM
HallsofIvy
Quote:

Originally Posted by decoy808
want to check a question and solution.

differentiate with respect to x

$\displaystyle y= (x^3+4)/x$

= $\displaystyle x^2+4x^-{^1}$
( this is the line im confused about why does the x^3 become x^2 and not 3x) or is it wrong

Yes, that is correct. It is $\displaystyle \frac{x^4+ 4}{x}= \frac{x^4}{x}+ \frac{4}{x}$. The "$\displaystyle x^3$" becomes "$\displaystyle x^2$" because the x in the denominator cancels one x in the numerator.

Quote:

$\displaystyle dy/dx = 2x-4x^{-2}$

many thanks
Yes, that derivative is correct.

Two minutes late again! I need to type faster!
• Feb 17th 2010, 04:41 AM
decoy808
how do i do y= (x+1)/x
• Feb 17th 2010, 02:54 PM
Prove It
Quote:

Originally Posted by decoy808
how do i do y= (x+1)/x

$\displaystyle \frac{x + 1}{x} = \frac{x}{x} + \frac{1}{x} = 1 + x^{-1}$.

So the derivative is $\displaystyle -1x^{-2} = -\frac{1}{x^2}$.