Showing that $f(x) = \frac{x}{10^{\frac{a}{10}x}-1}, a > 0$ is always decreasing

Question

I'd like to show that the following function is always decreasing: $$f(x) = \frac{x}{10^{\frac{a}{10}x}-1}, a > 0$$ that is $\frac{df(x)}{dx}<0,\forall x$.

I have found the first derivative and limited the reasoning to the numerator only since the denominator is squared it is always greater than zero (I didn't consider the study of possible zero value).

So I need to show that the numerator is always less or equal to zero.

Here the formula I have in front of me, I've tried several ways to show that but, without success.

$$(1-\frac{a}{10}x\ln(10))10^{\frac{a}{10}x}-1 \leq 0$$

Could you please give me a hint?

Thanks in advance

Regards

Answer 1

Hint: By the Quotient rule i got this here $$f'(x)=-1/10\,{\frac {x{10}^{1/10\,ax}a\ln \left( 10 \right) -10\,{10}^{1/10 \,ax}+10}{ \left( {10}^{1/10\,ax}-1 \right) ^{2}}} $$ We must know something about the $a$

Answer 2

If we prove that $$f(u)=(1-u)e^{u} \leq 1$$

we are done, because your numerator can be rewritten as $(1-kx)e^{kx} - 1$ with $k >0$.

Taking the derivative we obtain $f'(u) = -ue^u$ which means that $f(u)$ has a maximum in $u=0$. But $f(0) = 1$ so $f(u) \leq 1$.