help with differentiating an expression!

[bayesrule]

Hi!

This is my first post. Thanks for creating the forum, Ted!

I am having trouble differentiating the following with respect to b1: =(V1-b1)b1/?
I am able to differentiate the same expression before /? is introduced, yielding the answer b1=v1/2

I see the answer is the same even when /? is introduced, but I don't seem to get there. I tried to use the product rule but I think where I'm going wrong is finding the derivative of b1/?. If it was ?/b1 the derivative would be -?b1[sup:23bftjju]-2[/sup:23bftjju], but I can't figure what it is the other way round!

Any help much appreciated.

Thank you

 

[mmm4444bot]

=(V1-b1)b1/?

b1 = v1/2

The equals sign in the first line above is a typographical error; I do not understand why the second line is an equation.

Also, you may not interchange upper and lower case letters for the same symbol.

In other words, the symbol v1 is not the same as V1.

Please use the [Preview] button, to proofread your posts.

I'm thinking that the derivative of the expression in the first line above (ignoring the equals sign) with respect to b1 is:

(V1 - 2*b1)/?

 

[bayesrule]

OK. Please excuse me. Let me type it as given.

?(b[sub:c3fz86o0]1[/sub:c3fz86o0], v[sub:c3fz86o0]1[/sub:c3fz86o0]) = (v[sub:c3fz86o0]1[/sub:c3fz86o0]-b[sub:c3fz86o0]1[/sub:c3fz86o0])b[sub:c3fz86o0]1[/sub:c3fz86o0]/?

Differentiating this, setting the derivative equal to zero and solving for b[sub:c3fz86o0]1[/sub:c3fz86o0] is how I am trying to arrive at b[sub:c3fz86o0]1[/sub:c3fz86o0]=v[sub:c3fz86o0]1[/sub:c3fz86o0]/2

Thanks

 

[bayesrule]

Re:

I'm thinking that the derivative of the expression in the first line above (ignoring the equals sign) with respect to b1 is:

(V1 - 2*b1)/?

Great, thanks. Could you tell me how you got there please? And how does this simplify to b[sub:2931t4k4]1[/sub:2931t4k4]=v[sub:2931t4k4]1[/sub:2931t4k4]/2 ?

As an equation (see previous post), you have 0=(V[sub:2931t4k4]1[/sub:2931t4k4] - 2*b[sub:2931t4k4]1[/sub:2931t4k4])/?, so do you just times both sides by /? to take it off?

Much appreciated

 

[galactus]

\(\displaystyle \frac{(v_{1}-b_{1})b_{1}}{B}\)

All are constants except b1:

Expand:

\(\displaystyle \frac{v_{1}b_{1}}{B}-\frac{b_{1}^{2}}{B}\)

Differentiate each term:

\(\displaystyle \frac{v_{1}}{B}-\frac{2b_{1}}{B}=0\)

\(\displaystyle \frac{v_{1}}{B}=\frac{2b_{1}}{B}\)

\(\displaystyle b_{1}=\frac{Bv_{1}}{2B}\)

The B's cancel and we are left with:

\(\displaystyle b_{1}=\frac{v_{1}}{2}\)