# derivative question

• Jan 19th 2010, 07:49 PM
stumped765
derivative question
Hi this should be easy but im having a bit of a brain block.

find the derivative of this with respect to x

-(x+1)(ln(y*z*r)
• Jan 19th 2010, 08:04 PM
Prove It
Quote:

Originally Posted by stumped765
Hi this should be easy but im having a bit of a brain block.

find the derivative of this with respect to x

-(x+1)(ln(y*z*r)

Are $y, z$ and $r$ constants or are any of them functions of $x$?
• Jan 19th 2010, 09:54 PM
stumped765
they can be regarded as constants
• Jan 19th 2010, 10:14 PM
Prove It
Quote:

Originally Posted by stumped765
Hi this should be easy but im having a bit of a brain block.

find the derivative of this with respect to x

-(x+1)(ln(y*z*r)

If $y, z, r$ are constants, then $\ln{(yzr)}$ is also a constant.

So $\frac{d}{dx}\left[-(x + 1)\ln{(yzr)}\right] = -\ln{(yzr)}$.
• Jan 20th 2010, 05:56 AM
HallsofIvy
Quote:

Originally Posted by Prove It
If $y, z, r$ are constants, then $\ln{(yzr)}$ is also a constant.

So $\frac{d}{dx}\left[-(x + 1)\ln{(yzr)}\right] = -x\ln{(yzr)}$.

Typo- that 'x' should not be in the final answer. This is simply -x C+ C (where I have written the constant ln(yzr) as "C"), linear in x, so its derivative is the constant -C= -ln(xzr).
• Jan 20th 2010, 10:25 PM
Prove It
Quote:

Originally Posted by HallsofIvy
Typo- that 'x' should not be in the final answer. This is simply -x C+ C (where I have written the constant ln(yzr) as "C"), linear in x, so its derivative is the constant -C= -ln(xzr).

Yes, typo. Thanks.