(Easy question) Problem with deriving unknown function

Question

in my lecture slides we maximize the following utility function for s: $$u(x-s) + u(y+s)$$ and receive $$-u'(x-s) + u'(y+s)$$

1. Let's just focus on the first part $u(x-s)$ that is derived to $-u'(x-s)$. Obviously the chain rule has been applied here.

2. Now lets say $u(x-s)=\sqrt {x-s}$. This can be derived to $-0.5(x-s)^{-0.5}$

3. So, if $u(x-s)=\sqrt {x-s}$ then $-u'(x-s)$ should be equal to $-0.5(x-s)^{-0.5}$. Which is obviously not the case because $-u'(x-s) \neq u'(x-s)$? or respectively $(-)-0.5(x-s)^{-0.5} \neq -0.5(x-s)^{-0.5}$

Where is my error?

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Probably not relevant here but just for completenes:

• $u'\gt0$ and $u''\lt0$
• $x$ is the income in period 1
• $y$ is the income in period 2
• $s$ is savings
• intertemporal preferences (whatever that means.. anyone knows what that means?)

Answer 1

3. So, if $u(x-s)=\sqrt {x-s}$ then $-u'(x-s)$ should be equal to $-0.5(x-s)^{-0.5}$. Which is obviously not the case because $-u'(x-s) \neq u'(x-s)$? or respectively $(-)-0.5(x-s)^{-0.5} \neq -0.5(x-s)^{-0.5}$

Where is my error?

If $u(x-s)=\sqrt {x-s}$ then

$$\left(u(x-s)\right)'\neq u'(x-s)$$

as letting $w=u-s$ we see that

$$\left(u(x-s)\right)'=\left(u(w)\right)'=u'(w)\frac{dw}{ds}=-u'(w)$$

while

$$u'(x-s)=u'(w)$$

The error that you made is that you forgot to apply the chain rule to the inside function which in your case is

$$\frac{d}{ds}(x-s)=-1$$

Answer 2

In your notation, $$u'(x-s) = \left. \frac{\partial}{\partial z} u(z) \right|_{z = x-s}$$

In other words, take the function $u=u(z)$ and compute its first derivative, then evaluate the derivative at $z=x-s$.

Answer 3

It is just the chain rule:

$(f(g(s)))' =g'(s)f'(g(s)) $.

If $g(s) = x-s$ then $g'(s) = -1$ so $(u(x-s))' =(x-s)'u'(x-s) =-u'(x-s) $ since $(x-s)' = -1$.

Remember, all derivatives are being taken with respect to $s$.