1. ## Help with Differentiation

How would I go about doing a problem like this?

Find dy/dx for y= x^(9/8)

Just one more thing, where can I find a list to all the commands for MATH tags?

2. Originally Posted by Zanderist
How would I go about doing a problem like this?

Find dy/dx for y= x^(9/8)
If $y = x^n$, where $n$ is a constant, then $\frac {dy}{dx} = nx^{n - 1}$

Just one more thing, where can I find a list to all the commands for MATH tags?

3. Originally Posted by Jhevon
If $y = x^n$, where $n$ is a constant, then $\frac {dy}{dx} = nx^{n - 1}$

My problem as it was $x^\frac{9}{8}$

$\frac {dy}{dx} = \frac {9x^\frac{9}{8}-1}{8}$

The "1" by the way is with the Exponent.

So with that said it should come out to be:

$\frac {9x^8}{8}$

4. Originally Posted by Zanderist
$\frac {dy}{dx} = \frac {9x^\frac{9}{8}-1}{8}$

The "1" by the way is with the Exponent.

So with that said it should come out to be:

$\frac {9x^8}{8}$
$\frac{9}{8}-1 = \frac{1}{8}$

5. y= x^(9/8)
iff
ln(y)=ln(x^(9/8))
ln(y)=(9/8)ln(x), property of logs.
(dy/dx)(1/y)=(9/8)(1/x)
(dy/dx)=y(9/8)(1/x)
(dy/dx)=(x^(9/8))(9/8)(1/x)

6. Originally Posted by Chris11
y= x^(9/8)
iff
ln(y)=ln(x^(9/8))
ln(y)=(9/8)ln(x), property of logs.
(dy/dx)(1/y)=(9/8)(1/x)
(dy/dx)=y(9/8)(1/x)
(dy/dx)=(x^(9/8))(9/8)(1/x)
and x^(9/8)(1/X)= x^(1/8) as before.

Do you always use logarithms to differentiate x^n?