Best way to differentiate this?

[moogle12]

[MATH]0.5/(0.5+c) (-0.5/(0.5+c) c+1-0.5/(0.5+c))+(1-0.5/(0.5+c) )(0.5(1-0.5/(0.5+c)))[/MATH]
I'm afraid I don't know LaTeX and don't have time to learn it for this one specific post. I'd like to find the first and second derivatives of this but I'm a little intimidated. How best should I go about it? I am aware of the chain, product and quotient rules but using these seems like I'd be incredibly prone to making mistakes due to how much there is going on here.

 

[Steven G]

The way you wrote the problem means it looks like this:

[math]\dfrac {0.5}{0.5+c}(\dfrac{-0.5c}{0.5+c} + 1 -\dfrac{0.5}{0.5+c}) + \dfrac{1-0.5}{0.5+c}*0.5*(1-\dfrac{0.5}{0.5+c})[/math]
Is this correct? I certainly would simplify first before computing the derivative.

 

[HallsofIvy]

The first thing I would do is let y= 0.5+ c. Then we have \(\displaystyle \frac{0.5}{y}\left(\frac{-0.5c}{y}+ 1- \frac{0.5}{y}\right)+ \frac{1- 0.5}{2y}\left(1- \frac{0.5}{y}\right)\).

I would then recognize that \(\displaystyle \frac{-0.5c}{y}- \frac{0.5}{y}= -0.5\frac{c+ 1}{y}= -0.5{y+0.5}{y}\) and that 1- 0.5= 0.5 so that can be written \(\displaystyle \frac{-0.25}{y}\left(\frac{y+ 0.5}{y}+ 1\right)+ \frac{1}{4y}\left(\frac{0.5}{y}\right)= \frac{-1}{4y}\left(2+ \frac{1}{2y}\right)+ \frac{1}{8y^2}= -\frac{1}{2y}- \frac{1}{8y^2}+ \frac{1}{8y^2}= -\frac{1}{2y}= -\frac{1}{2}y^{-1} \).

And, of course, since y= 0.5+ c, \(\displaystyle \frac{dy}{dc}= 1\).

 

[Dr.Peterson]

[MATH]0.5/(0.5+c) (-0.5/(0.5+c) c+1-0.5/(0.5+c))+(1-0.5/(0.5+c) )(0.5(1-0.5/(0.5+c)))[/MATH]
I'm afraid I don't know LaTeX and don't have time to learn it for this one specific post. I'd like to find the first and second derivatives of this but I'm a little intimidated. How best should I go about it? I am aware of the chain, product and quotient rules but using these seems like I'd be incredibly prone to making mistakes due to how much there is going on here.

You can get around the hardest parts of LaTeX by using backquotes instead of MATH /MATH around what you wrote, which treats it as AsciiMath. Here's the result, without my changing anything else:

`0.5/(0.5+c) (-0.5/(0.5+c) c+1-0.5/(0.5+c))+(1-0.5/(0.5+c) )(0.5(1-0.5/(0.5+c)))`​

Please respond to the request to verify that what you wrote is what you mean. Then show an attempt to simplify it (perhaps as an image of your handwritten work) so we can check it out. But I absolutely agree that the algebra to simplify first will be much easier than the differentiation rules. There are lots of common denominators you can use, for example.

 

[Steven G]

By the way you never said what you are taking the derivative with respect to. If it is not wrt c, then the derivative is 0.