Partial derivative

**el123** · September 18th 2009, 06:30 PM

$3x^2 \ln(1-xy)$

the partial derivatvie with respect to y is

$\frac{x}{1-xy}$

is that right?

**Prove It** · September 18th 2009, 06:35 PM

Originally Posted by el123

$3x^2 \ln(1-xy)$

the partial derivatvie with respect to y is

$\frac{x}{1-xy}$

is that right?

Not quite.

It's actually

$3x^2\left(\frac{-x}{1 - xy}\right)$

$= -\frac{3x^3}{1 - xy}$ .

**el123** · September 18th 2009, 06:39 PM

but i thought you treat x as a constant so its derivative is 0?

**Prove It** · September 18th 2009, 06:44 PM

Originally Posted by el123

but i thought you treat x as a constant so its derivative is 0?

Yes, it is treated like a constant.

What do you do to differentiate a constant times a function?

It's the constant times the derivative of that function.

i.e. $\frac{d}{dx}\left[c\cdot f(x)\right] = c\cdot f'(x)$ .

So in your case, you have $3x^2 \ln{(1 - xy)}$

The $3x^2$ is your constant.

So the partial derivative will be

$3x^2 \frac{\partial}{\partial y}\left[\ln{(1 - xy)}\right]$ .

**el123** · September 18th 2009, 06:46 PM

true to that , thank you very much !

Thread: Partial derivative

LinkBack

Thread Tools

Display

Partial derivative

Similar Math Help Forum Discussions

Please help with partial derivative.

Derivative of arctan in a partial derivative

another partial derivative

help me!!! (partial derivative)

Partial Derivative

Search Tags