Easy way to determine the sign of derivation

Question

I was wondering if there is another way to determine the sign of the derivative of the function $f(x)$. Sometimes it is quite difficult (and lengthy) to derive the derivation for $\partial f/ \partial x$.

I just need a qualitative answer, i. e. I just want to know the sign of the derivation, not a quantitative value.

I edited my post (14.11.) and tried to derive a new proof:

We have a function with the following form:

$$f(x)= {{(1+g(x))^T-(1+g(x))^{t-t_0}} \over {(1+g(x))^T-1} } $$

and we know the following things:

${\partial g(x) \over \partial x} <0$
$T>t>t_0>0$ and are non-negative integers.

Proposition: $f(x)$ is decreasing with x

(This is of course only true for the additional statements above for $g(x)$ and $T, t, t_0$)

Proof:

For the proof we use the following logic: The decrease in the nominator by a small increase in $x$ must be smaller than the decrease in the denominator. Otherwise the fraction would not be decreasing with an increase in $x$ and the proposition would not be true. If for example the decrease in the nominator would be larger than the decrease in the denominator the fraction would increase and the proposition can not be true.

For that statement to be true the following inequality for a small increase in $x$ denoted by $\Delta$ must always hold:

$$((1+g(x))^T-(1+g(x))^{t-t_0})-(1+g(x+\Delta))^T+(1+g(x+\Delta))^{t-t_0}<((1+g(x))^T-1)- ((1+g(x+\Delta))^T-1)$$

Reducing the inequality we can see that:

$$-(1+g(x))^{t-t_0})+(1+g(x+\Delta))^{t-t_0}<0$$ which must be true all the time because $g(x)>g(x+\Delta)$ (see the statement above for ${\partial g(x) \over \partial x} <0$

Answer 1

I am afraid there is no easy way to show this for general $t,t_0,T$ and $g$. But here is a (long) way to show that $f$ is always decreasing for the assumptions you have given.

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Let's define $h(x):=1+g(x)$, $\alpha:=-t+t_0+T$ and $\beta:=T.$ So we can work with this easier looking expression:

$$ f(x)= \frac{[h(x)]^T-[h(x)]^{t-t_0}}{[h(x)]^T-1}= \frac{1-[h(x)]^{t-t_0-T}}{1-[h(x)]^{-T}}= \frac{1-[h(x)]^{-\alpha}}{1-[h(x)]^{-\beta}}=:\frac{u(x)}{v(x)}. $$

Note that we have $\beta>\alpha>0$, $h(x)>1$ and $h'(x)<0$. We can simply apply the quotient rule to find

$$f'(x)=\frac{u'v-v'u}{v^2}<0\qquad\text{if and only of}\qquad u'v<v'u.$$

and because $u,v>0$ and $u',v'<0$, this is equivalent to ask for $f(x)=u/v<u'/v'$. In the case of above function this results in the claim

$$f(x)<\frac{\alpha [h(x)]^{-\alpha-1}\cdot h'(x)}{\beta [h(x)]^{-\beta-1}\cdot h'(x)}=\frac\alpha\beta [h(x)]^{\beta-\alpha}.$$

This means that $f$ is decreasing in some $x$ if and only if this holds true at this specific point. So in order to prove that $f$ is always decreasing, we need to show that for all real numbers $h>1$ and $\beta>\alpha>0$ we have

$$ \frac{1-h^{-\alpha}}{1-h^{-\beta}}<\frac\alpha\beta h^{\beta-\alpha}\quad\implies\quad\frac1\alpha(h^\alpha-1)<\frac1\beta(h^\beta-1). $$

This means we are asked to show that $\gamma(x)=(h^x-1)/x$ is always increasing on $(0,\infty)$ for $h>1$. Here again we need some dirivative magic:

$$\gamma'(x)=\frac{h^x\log h\cdot x-h^x+1}{x^2}>0\qquad\text{if and only if}\quad h^{-x}>1-x\log(h).$$

But this condition is always true since $h^{-x}$ is a convex function and $1-x\log(h)$ is its tangent at $x=0$. Convex functions always dominate their tangents.