chain rule problem: u(c) = -(y/2)(c-z) the (c-z) is squared

[ollessendro]

Hi everyone, I'm having problems with differentiation.

Its on the following .........

u(c) = -(y/2)(c-z) the (c-z) is squared

I have the answer as u'(c) = (c-z)y

I just cant see why?! Can anyone help me?

 

[jwpaine]

\(\displaystyle \L u(c) = -\frac{y}{2}(c-z)^{2}\)

Use the product rule and chain rule:

\(\displaystyle \L u'(c) = \frac{d}{dc}[-\frac{y}{2}](c-z)^{2}\,+\,-\frac{y}{2}\cdot \frac{d}{dc}[(c-z)^{2}]\)

\(\displaystyle \L u'(c) = [0](c-z)^{2}\,+\,-\frac{y}{2}\cdot 2(c-z)\frac{d}{dc}[(c-z)]\)

\(\displaystyle \L u'(c) = 0+\,-\frac{y}{2}\cdot 2(c-z)[1]\)

\(\displaystyle \L u'(c) = -\frac{y}{2}\cdot 2(c-z)\)

\(\displaystyle \L u'(c) = -y(c-z)\)

Although I got -y(c-z) as an answer........ maybe someone else can check to see if I did everything correctly... I haven't been doing this stuff very long

John

 

[Deleted member 4993]

Re: chain rule problems

I'm having problems with differentiation.

Its on the following .........

u(c) = -(y/2)(c-z) the (c-z) is squared

I have the answer as u'(c) = (c-z)y

I just cant see why?! Can anyone help me?

In this case, y and z are treated as constants - because 'u' has been defined to be function of 'c' only.

as written

\(\displaystyle u(c) = -\frac{y}{2}\cdot\(c-z)^2\)

u'(c) = - y/2 * 2 * (c-z) = - y * (c-z) <---- which is not same as the answer you posted.

 

[Deleted member 4993]

\(\displaystyle \L u(c) = -\frac{y}{2}(c-z)^{2}\)

Use the product rule and chain rule:

\(\displaystyle \L u'(c) = \frac{d}{dc}[-\frac{y}{2}](c-z)^{2}\,+\,-\frac{y}{2}\cdot \frac{d}{dc}[(c-z)^{2}]\)

\(\displaystyle \L u'(c) = -[\frac{1}{2}](c-z)^{2}\,John - how did you get (1/2) here!+\,-\frac{y}{2}\cdot 2(c-z)\frac{d}{dc}[(c-z)]\)

\(\displaystyle \L u'(c) = -[0](c-z)^{2}\,+\,-\frac{y}{2}\cdot 2(c-z)[1]\)

\(\displaystyle \L u'(c) = -\frac{y}{2}\cdot 2(c-z)\)

\(\displaystyle \L u'(c) = -y(c-z)\)

Although I got -y(c-z) as an answer........ maybe someone else can check to see if I did everything correctly... I haven't been doing this stuff very long

John

 

[ollessendro]

Thanks John.

That's the answer I keep getting.

Perhaps there is a typo in the question I have ......... and there is no minus sign in front. Then I can get the answer.

 

[jwpaine]

\(\displaystyle \L u(c) = -\frac{y}{2}(c-z)^{2}\)

Use the product rule and chain rule:

\(\displaystyle \L u'(c) = \frac{d}{dc}[-\frac{y}{2}](c-z)^{2}\,+\,-\frac{y}{2}\cdot \frac{d}{dc}[(c-z)^{2}]\)

\(\displaystyle \L u'(c) = -[\frac{1}{2}](c-z)^{2}\,John - how did you get (1/2) here!+\,-\frac{y}{2}\cdot 2(c-z)\frac{d}{dc}[(c-z)]\)

\(\displaystyle \L u'(c) = -[0](c-z)^{2}\,+\,-\frac{y}{2}\cdot 2(c-z)[1]\)

\(\displaystyle \L u'(c) = -\frac{y}{2}\cdot 2(c-z)\)

\(\displaystyle \L u'(c) = -y(c-z)\)

Although I got -y(c-z) as an answer........ maybe someone else can check to see if I did everything correctly... I haven't been doing this stuff very long

John

That was a typo I forgot to delete in my latex.

 

[Deleted member 4993]

When you are dealing with "multiplied by constant" - you don't need to use product rule - simply take the constant outside your "domain" of differentiation.

 

[jwpaine]

I was thinking I had to use the product rule, because \(\displaystyle \L\frac{d}{dx}(uv) = [\frac{d}{dx}u]v\,\,+\,\,u[\frac{d}{dx}v]\)

I figured for the posted problem, that u was a function, not a constant, where u = \(\displaystyle \L -\frac{y}{2}\) but sense we are differentiating with respect to c, than \(\displaystyle \L -\frac{y}{2}\) can be treated as a constant?