How do I differentiate the term $\ln((1+\beta)^x-1)$ with respect to $x$?
Is it possible to do it this way:
$$\frac{1}{(1+\beta)^x-1}$$
But i get stuck if i do the normal differentiation.
$\endgroup$ 14 Answers
$\begingroup$You have to multiply your expression by the derivative of $(1+\beta)^x$. Note $(1+\beta)^x=\exp(x\log (1+\beta))$...
$\endgroup$ $\begingroup$Let $y=\ln ((1+ \beta)^x-1)), e^y= (1+ \beta)^x-1 \\ e^y+1=(1+ \beta)^x \\ \ln (e^y+1)=x \ln (1+ \beta)$.
Hopefully you can see where to go from here.
$\endgroup$ $\begingroup$Implicit differentiation might be needed.
Hint: Use the chain rule: $$(\ln u)'=u'/u$$
$\endgroup$ $\begingroup$\begin{align} & \frac d {dx} \ln ((1+\beta)^x - 1) \\[10pt] = {} & \frac{1}{(1+\beta)^x - 1} \cdot\frac d {dx}((1+\beta)^x-1) \tag{chain rule} \\[10pt] = {} & \frac{1}{(1+\beta)^x - 1} \cdot(1+\beta)^x\cdot\ln(1+\beta). \end{align}
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